Tom's normal working hours on Monday are 7 hours, for which he will be paid £15 x 7 = £105.Tom worked for an additional 4 hours on Monday, for which he will receive overtime pay. His overtime rate is 1 x £15 = £15 per hour.So, Tom's overtime earnings for Monday are equal to £15 x 4 = £60.Therefore, the total amount Tom earned on Monday is £105 + £60 = £165.Total = £165
[25] 1251 3) Fit the data given in the table of problem (2) to an exponential equation of the form y = 1 + aebx by linearizing the equation and using linear regression to determine the coefficients a and b. Use this result to estimate the value of y at x =
Using the exponential regression feature of the calculator to find the equation of the regression line, we get that [tex]$$y = 0.8996 e^{1.3759x}.$$[/tex]
Given data, $$\begin{array}{|c|c|} \hline x & y\\ \hline 1 & 2.20\\ 2 & 3.60\\ 3 & 5.90\\ 4 & 9.70\\ 5 & 15.90\\ 6 & 26.00\\ \hline \end{array}.$$
The equation of the form is y = 1 + aebx;
Thus, the required equation is [tex]$$y = 1 + 0.8996 e^{1.3759x}.$$[/tex]
Finally, putting x = 7, we get
[tex]$$y = 1 + 0.8996 e^{1.3759(7)} \approx 156.76.$$[/tex]
Thus, the required equation is[tex]$$y = 1 + 0.8996 e^{1.3759x}.$$[/tex]Finally, putting x = 7, we get
[tex]$$y = 1 + 0.8996 e^{1.3759(7)} \approx 156.76.$$[/tex]
So, the value of y at x = 7 is approximately 156.76.
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For each of the following functions, obtain first AND second derivatives, expressed as f′(x) and f′′(x). A. f(x)=8x3+2x2−5x+3 B. f(x)=cos(4x2) C. f(x)=ex^5
Answers of derivative:
[tex]A \( f'(x) = 24x^2 + 4x - 5 \) \\and \( f''(x) = 48x + 4 \)\\\\B \( f'(x) = -8x \sin(4x^2) \) and \( f''(x)\\ = -8 \sin(4x^2) - 64x^2 \cos(4x^2) \)\\\\C \( f'(x) = 5x^4 e^{x^5} \) and \( f''(x) \\= 20x^3 e^{x^5} + 25x^8 e^{x^5} \)[/tex]
A. To find the first and second derivatives of[tex]\( f(x) = 8x^3 + 2x^2 - 5x + 3 \):\\First derivative:\( f'(x) = 24x^2 + 4x - 5 \)\\Second derivative:\( f''(x) = 48x + 4 \)\\\\B To find the first and second derivatives of \( f(x) = \cos(4x^2) \):\\First derivative:\( f'(x) = -8x \sin(4x^2) \)\\Second derivative:\( f''(x) = -8 \sin(4x^2) - 64x^2 \cos(4x^2) \)\\\\C. To find the first and second derivatives of \( f(x) = e^{x^5} \):\\First derivative:\( f'(x) = 5x^4 e^{x^5} \)\\Second derivative:\( f''(x) = 20x^3 e^{x^5} + 25x^8 e^{x^5} \)[/tex]
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Define R as the region bounded above by the graph of f(x)=x^2
and below by the x-axis over the interval [0,1]
Find the volume of the solid of revolution formed by revolving R
around the line x=−2.
The volume of the solid of revolution formed by revolving the region R, bounded above by the graph of [tex]\(f(x) = x^2\)[/tex] and below by the x-axis over the interval [0, 1], around the line [tex]\(x = -2\), is \(\frac{265\pi}{6}\)[/tex] cubic units.
To find the volume, we can use the method of cylindrical shells. The radius of each shell is the distance from the line [tex]\(x = -2\)[/tex] to the x-axis, which is 2 units. The height of each shell is the difference in the y-coordinates between the graph of [tex]\(f(x) = x^2\)[/tex] and the x-axis. Since the interval is from 0 to 1, the height of each shell varies from 0 to 1. The volume of each shell is given by the formula [tex]\(2\pi \cdot \text{{radius}} \cdot \text{{height}}\)[/tex].
Integrating this formula over the interval [0, 1] gives us the total volume of the solid of revolution. Evaluating the integral [tex]\(\int_{0}^{1} 2\pi \cdot 2 \cdot x^2 \,dx\)[/tex] gives us [tex]\(\frac{265\pi}{6}\)[/tex] cubic units.
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The equation x2+y2=4y can be converted to a. r=4secθ b. r=2cosθ c. r=4sinθ d. r=4sinθcosθ
Thus, the option is (c) r = 4sin(θ).
The equation x2 + y2 = 4y can be converted to an equation in polar coordinates as r = 4sin(θ).
Here's why:
We know that x = rcos(θ) and
y = rsin(θ).
So, x2 + y2 = r2cos2(θ) + r2sin2(θ)
= r2(cos2(θ) + sin2(θ))
= r2.
This means that r2 = 4y, which implies that r = 2√y.
Substituting this value of r into the equation x2 + y2 = 4y, we get x2 + (r2/4) = r, which is the equation in polar coordinates.
So, we have r = 2√y, which can be simplified as r = 4sin(θ).
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Identify The Surface 2x² + X² = 4x + 2y² Through Completing The Square.
To identify the surface given by the equation 2x² + x² = 4x + 2y² through completing the square, we can rewrite the equation in a more convenient form.
Let's start by rearranging the equation: 3x² - 4x + 2y² = 0. To complete the square for the x-terms, we focus on the quadratic part, 3x² - 4x. We divide the coefficient of x by 2 and square it: (-4/2)² = 4. Adding and subtracting this value inside the parentheses, we have 3x² - 4x + 4 - 4 + 2y² = 0. Rearranging the terms, we get (3x² - 4x + 4) + 2y² - 4 = 0. Now, we can rewrite the equation as (3x² - 4x + 4) + 2(y² - 2) = 0. Simplifying further, we have 3(x - 2/3)² + 2(y - √2)² = 8/3. The equation is now in the form (x - h)²/a² + (y - k)²/b² = 1, which represents an ellipse. The center of the ellipse is at (2/3, √2), with the x-axis semi-major axis equal to √(8/9) and the y-axis semi-minor axis equal to √(4/3).
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if √ x √ y = 7 x y=7 and y ( 16 ) = 9 y(16)=9 , find y ' ( 16 ) y′(16) by implicit differentiation.
16(9) = 7 144 = 7Since this equation is not trueTo find y'(16) using implicit differentiation, we'll differentiate both sides of the equation with respect to x and solve for y'.
Given:
√x √y = 7
xy = 7
y(16) = 9
Let's differentiate both sides of the equation √x √y = 7 with respect to x using the chain rule:
d/dx (√x √y) = d/dx (7)
Using the chain rule:
(1/2) √y + (1/2√x)(dy/dx) = 0
Now let's differentiate both sides of the equation xy = 7 with respect to x:
d/dx (xy) = d/dx (7)
Using the product rule:
y + x(dy/dx) = 0
We are given y(16) = 9, which means when x = 16, y = 9.
Substituting x = 16 and y = 9 into the equation xy = 7:
16(9) = 7
144 = 7
Since this equation is not true .
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Section A: Answer any 2 questions from Q1 to Q3 Q1. (a) An office with dimensions of 20 m (L) x 15 m (W) x 4 m (H) has 50 staff. A ventilation system supplying outdoor air to this office at a designed flow rate of 10 L/s/person. The outdoor CO₂ concentration is 300 ppm. The initial concentration of CO₂ in the office is 350 ppm and the CO₂ emission rate from each person is 0.01 L/s respectively. (i) Determine the CO₂ concentration in ppm in the office at the end of the first 3 hours if it is full house. (10 Marks) (ii) Determine the steady state concentration of CO2 in ppm in the office at full house condition. (4 Marks) (iii) Comment on whether any IAQ class can be achieved in the office in accordance with the IAQ Certification Scheme by Environmental Protection Department (EPD). (2 Marks) (iv) If the office is to achieve Excellent Class at steady state in accordance with the IAQ Certificate Scheme, determine the minimum flow rate of outdoor air (in m³ per hour) required for the ventilation system. (3 Marks) Given: C(t)=Coe^-qv 1/v + CvQv+G/Qv (1-e^-Qv 1/v) (b) Renovation works will be carried out for a customer services centre in a high-rise commercial building at Causeway Bay. The renovation contractor is required to implement necessary preventive measures to reduce the environmental nuisance to the adjacent office units on the same floor. Propose FOUR suitable measures to minimize IAQ problems during renovation works. (6 Marks)
The minimum flow rate of outdoor air required is approximately 14400 m³/h. And we can achieve the desired IAQ class depending on the CO2 concentration.
(i) CO2 concentration after 3 hours if it is full house:
To find the CO2 concentration after 3 hours, we need to use the following formula:
[tex]C(t)=Coe^-qv 1/v + CvQv+G/Qv (1-e^-Qv 1/v)[/tex]
Here, C(t) is the CO2 concentration after time t, Co is the initial concentration, v is the volume of the room, q is the CO2 emission rate per person, Qv is the outdoor airflow rate, and G is the CO2 generation rate per unit time.
We are given that the initial CO2 concentration is 350 ppm, the outdoor CO2 concentration is 300 ppm, the CO2 emission rate from each person is 0.01 L/s, and the designed outdoor airflow rate is 10 L/s/person.
So, we have:v = 20 × 15 × 4 = 1200 m³q = 0.01 L/s/person × 50 person = 0.5 L/sQ
v = 10 L/s/person × 50 person = 500 L/sG = 0 (because no other CO2 source is mentioned)
Therefore,
C(3 hours) = [tex]350e^-500×3×3600/1200 + 300×500/1200 + 0.5/1200 (1-e^-500/1200×3×3600)[/tex])≈ 1023 ppm
So, the CO2 concentration in the office at the end of the first 3 hours if it is full house is approximately 1023 ppm.
(ii) Steady-state CO2 concentration at full house condition:To find the steady-state CO2 concentration at full house condition, we need to equate the CO2 generation rate to the outdoor airflow rate times the concentration difference:
Qv(C - C0) = G
Here, C is the steady-state CO2 concentration, and C0, Qv, and G are as given above.
So,C = G/Qv + C0 = 0.5 L/s ÷ 500 L/s × 3600 s/h + 300 ppm ≈ 303 ppm
So, the steady-state concentration of CO2 in ppm in the office at full house condition is approximately 303 ppm.
(iii) IAQ class that can be achieved in the office:
The IAQ Certification Scheme by Environmental Protection Department (EPD) requires that the CO2 concentration in an occupied space should not exceed 1000 ppm.
Since the CO2 concentration in the office is less than 1000 ppm at steady-state, any IAQ class can be achieved in the office.
(iv) Minimum flow rate of outdoor air to achieve Excellent Class:
To achieve Excellent Class, the steady-state CO2 concentration must not exceed 800 ppm. So, we need to solve the equation
Qv(800 - 350) = 0.5 L/s
to find the minimum flow rate of outdoor air required:
Qv = 0.5 L/s ÷ 450 ppm × 3600 s/h ≈ 4 m³/s or 14400 m³/h
These results show the importance of proper ventilation in maintaining good indoor air quality, particularly in densely occupied spaces like offices.
Thus, we can conclude that to maintain indoor air quality, proper ventilation plays a very important role. In case of densely occupied places like offices, the minimum flow rate of outdoor air required is approximately 14400 m³/h. And we can achieve the desired IAQ class depending on the CO2 concentration.
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Triangle X Y Z is shown. Angle X Y Z is a right angle and angles Y Z X and Z X Y are 45 degrees. The length of side Y X is 9 centimeters.
Triangle XYZ is a right triangle because it has a right angle. It also has angles of 45 degrees, 45 degrees, and 90 degrees, making it a right triangle with two identical sides. In this case, the side YZ measures 9 cm, not BC.
We've done that.
The triangle X Y Z is shown. Angles Y Z X and Z X Y are 45 degrees apart, but angle X Y Z is a straight angle. The side YX is 9 centimeters long.
Segment XY measures 9 cm in length.
We must assess whether the claims made about triangle XYZ are true.
A triangle with a right angle or two perpendicular sides is referred to as a right triangle, right-angled triangle, orthogonal triangle, or more technically, a rectangular triangle.
The right angle with angles of 45, 45, and 90 has two identical sides.
The right angle's side YZ=9 cm BC as a result.
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The correct question would be as
Triangle X Y Z is shown. Angle X Y Z is a right angle and angles Y Z X and Z X Y are 45 degrees. The length of side Y X is 9 centimeters.
The length of segment XY is 9 cm. Which statements regarding triangle XYZ are correct? Select two options.
YZ = 9 cm
XZ = 9 cm
XZ = 9 StartRoot 2 EndRoot cm
XZ = 2(XY)
YZ is the longest segment in △XYZ.
Find the first five nonzero terms of the Maclaurin expansion of f(x)=e −x
+cos(x). (Use symbolic notation and fractions where needed.)
We can use the Taylor series expansion for each term separately and then combine them to find the Maclaurin expansion of the function f(x) = e(-x) + cos(x).
To determine the first nonzero terms using maclaurin series
The Maclaurin series expansion for e^(-x) is given by:
[tex]e^(-x) = 1 - x + (x^2)/2! - (x^3)/3! + (x^4)/4! - ...[/tex]
The Maclaurin series expansion for cos(x) is given by:
[tex]cos(x) = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + (x^8)/8! - ...[/tex]
To find the terms of the combined series, we simply add the corresponding terms from both expansions. Let's calculate the first five nonzero terms:
Term 1: 1 [tex](from e^(-x))[/tex]
Term 2: -[tex]x (from e^(-x))[/tex]
Term 3:[tex](x^2)/2! - (x^2)/2! = 0[/tex] (canceling terms)
Term 4: [tex]-(x^3)/3! (from e^(-x))[/tex]
Term 5: [tex](x^4)/4! - (x^4)/4! = 0[/tex] (canceling terms)
Therefore, the first five nonzero terms of the Maclaurin expansion of [tex]f(x) = e^(-x) + cos(x)[/tex] are:
[tex]f(x) = 1 - x - (x^3)/3! + ...[/tex]
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A hollow steel ball of mass 1 kg is suspended from a spring. This stretches the spring 0.6125 m. Starting at t=0, an external force equal to f(t)=8sin4t is applied to the system. The air resistance (in N ) of the moving ball numerically equals 8 times its "velocity (in meters per second). The ball is started in motion from the equilibrium position. Let u(t) be the displacement of the mass from equilibrium. Suppose that after t seconds the ball is u meters below its rest position. Find u (in meters) in terms of t. (Note that the positive direction is down.) Take as the gravitational acceleration 9.8 meters per second per second How would you characterize the damping in this mass-spring system?
The overdamped system has two real, distinct roots when a quadratic equation is formed.The expression for displacement can be written as:u(t) = (16/17) - (1/17) e^(-25t) sin(17t), for 0 ≤ t < π/2.The value of u (in meters) in terms of t is given by the expression:(16/17) - (1/17) e^(-25t) sin(17t)
The damping in this mass-spring system can be characterized as overdamped.What is damping?The process of reducing the amplitude of oscillations of a system is known as damping. In a system subjected to any oscillatory force, damping is an energy-dissipative mechanism. The amount of damping in a system decides the type of system, such as underdamped, overdamped, or critically damped.What is overdamping?When the damping force is greater than the critical damping value, a system is considered to be overdamped. The damping force resists the system's motion, causing it to move more gradually to its equilibrium position without any oscillations. The system becomes too slow to reach equilibrium as the damping increases, causing the system's motion to become slow and sluggish. The overdamped system has two real, distinct roots when a quadratic equation is formed.The expression for displacement can be written as:u(t)
= (16/17) - (1/17) e^(-25t) sin(17t), for 0 ≤ t < π/2.The value of u (in meters) in terms of t is given by the expression:(16/17) - (1/17) e^(-25t) sin(17t)
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(i) Using methods of differentiation, locate and identify all stationary points,
and intercepts with x- and y-axis for y=4-32 +6.
(ii) Sketch the graph of the function, y = 4 -3x²+6, using the information obtained from part (i) on a piece of graph paper. All the stationary point(s) and axes intercept(s) must be clearly indicated and shown on the graph
To locate and identify all stationary points and intercepts of the function y = 4 - 3x² + 6, we differentiate the function with respect to x to find its derivative. The stationary points occur where the derivative is equal to zero or undefined. By setting the derivative equal to zero and solving for x, we can find the x-coordinates of the stationary points. The y-coordinates can be found by substituting these x-values back into the original function. The intercepts with the x-axis occur when y = 0, and the intercept with the y-axis occurs when x = 0. We can then sketch the graph using this information.
To find the stationary points, we differentiate the function y = 4 - 3x² + 6 with respect to x. The derivative of y with respect to x, dy/dx, is -6x. Setting dy/dx equal to zero, we have -6x = 0. Solving for x, we find x = 0. Substituting x = 0 into the original function, we get y = 4 - 3(0)² + 6 = 10. Therefore, the only stationary point is (0, 10).
To find the intercepts with the x-axis, we set y = 0 and solve for x. In this case, we have 4 - 3x² + 6 = 0. Simplifying the equation, we get 3x² = 10, and solving for x, we find x = ±√(10/3). These are the x-coordinates of the intercepts with the x-axis.
The intercept with the y-axis occurs when x = 0. Substituting x = 0 into the original function, we get y = 4 - 3(0)² + 6 = 10. Therefore, the intercept with the y-axis is at (0, 10).
Using this information, we can sketch the graph of the function y = 4 - 3x² + 6 on a piece of graph paper. We plot the stationary point (0, 10) and the x-intercepts (√(10/3), 0) and (-√(10/3), 0). We also plot the y-intercept (0, 10). The shape of the graph can be determined by the concavity of the function, which can be found by analyzing the second derivative. The graph should pass through these points and follow the concavity indicated by the second derivative.
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Find a vector function r(t) for the following curves. (a) line segment from P(1,−3,10) to Q(0,7,−1) (b) part of parabola y=−2x2 from (−1,−2) to (2,−8) (c) part of a circle of radius 4 , centered at the origin, traced out counterclockwise with initial point ( 4.0) and terminal point (0,−4)
Given : P(1, −3, 10) to Q(0, 7, −1), part of parabola y=−2x2 from (−1,−2) to (2,−8), and part of a circle of radius 4, centered at the origin, traced out counterclockwise with initial point ( 4.0) and terminal point (0,−4).a) We know that the equation of a line segment is r(t) = P + t(Q - P)Here, P(1, −3, 10) and Q(0, 7, −1)
Therefore, the vector function is given by:r(t) = (1 - t) i - 3(1 - t) j + 10(1 - t) k + 0 i + 7t j - t k= i - 3 j + 10 k + t(-i + 7 j - k)So, the vector function is r(t) = < i - t, -3 + 7t, 10 - t >.b) The part of the parabola y = -2x² from (-1, -2) to (2, -8) can be parameterized by x(t) = t and y(t) = -2t²
Here, x varies from -1 to 2 so we have the limits of integration as -1 to 2
Therefore, the vector function is given by :r(t) = i t + j (-2t²) = < t, -2t² >c) Part of a circle of radius 4, centered at the origin, traced out counterclockwise with initial point (4.0) and terminal point (0, −4). We know that the equation of a circle with radius r and center (a, b) is given by :r² = (x - a)² + (y - b)²
Here, the radius is 4 and the center is (0, 0)The equation is x² + y² = 16Using parametric equations x = r cosθ and y = r sinθ and taking θ = 0 at (4, 0), we have:r cosθ = 4 cosθ = 4/rr sinθ = 0 sinθ = 0
Here, θ varies from 0 to π/2We have the vector function :r(θ) = 4 cosθ i + 4 sinθ j = < 4 cosθ, 4 sinθ >Therefore, the vector function is r(θ) = < 4 cosθ, 4 sinθ >.
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ty(7 - y) Consider the initial value problem y' y(0) = yo. 7 a) Determine how the behavior of the solution as t increases depends on the initial value yo. If yo > 0, then If yo < 0, then If yo = 0, then b) Suppose that yo = 0.5. Find the time T at which the solution first reaches the value 6.98. NOTE: Round your answer to five decimal places. T Solve the initial value problem 4 cos(4x) y' ,y(0) = -1 3 + 2y and determine where the solution attains its maximum value (for 0 < x < 0.06317). - NOTE: Enclose arguments of functions in parentheses. For example, sin(2x). y(x) = = Find the value of x at which the solution attains a maximum. NOTE: Enter the exact answer. X =
(a)If y0=0, then the solution is identically 0.(b)Therefore, the required time T is approximately 4.31371 to five decimal places. (c) Therefore, we need to solve the equation y′=0 for 0
(a)The given initial value problem is y′=y(7−y),y(0)=y0.The behavior of the solution as t increases depends on the initial value y0 as follows: If y0>7, then the solution is increasing and unbounded.
If 00, then the solution is increasing and unbounded. If y0=7, then the solution is identically 7. If y0<0, then the solution is decreasing and approaches 0. If y0=0, then the solution is identically 0.
(b)The given initial value problem is y′=y(7−y),y(0)=0.5.
We need to find the time T at which the solution first reaches the value 6.98.Rearranging the given differential equation, we havey′7−y=y Integrating both sides with respect to t, we get∫dy7−y=y0∫dt+Cln|7−y|=-t+C1ln|7−y|=C2−t
Given that y(0)=0.5, we getln|7−0.5|=C2, which implies C2=ln(13.5).Hence, the above equation becomesln|7−y|=ln(13.5)−t.We need to solve for y when t=T and y=6.98.
Hence,6.98=7−e^(ln(13.5)−T) ⇒ e^(ln(13.5)−T)=0.02 ⇒ ln(13.5)−T=ln(0.02) ⇒ T=ln(13.5)−ln(0.02) ≈ 4.31371. Therefore, the required time T is approximately 4.31371 to five decimal places.
(c)The given initial value problem is 4cos(4x)y′=3+2y,y(0)=−1.Integrating the given differential equation, we get4∫cos(4x)dy=∫(3+2y)dx4sin(4x)+C=y
Solving the above equation for y, we gety=(4sin(4x)+C)/2−3/2We know that the solution is maximized when y′=0. Therefore, we need to solve the equation y′=0 for 0
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If all the graphs above have equations with form \( y=a b^{x} \), Which graph has the largest value for \( b \) ? Which graph has the smallest value for \( b \) ? Which graph has the largest value for
The graph with the smallest value for \( b \) will have the shallowest slope. The graph with the largest value for \( a \) will have the highest y-intercept.
To determine which graph has the largest value for \( b \), we need to compare the exponential growth rates.
Recall that in the equation \( y = ab^x \), the value of \( b \) determines the rate of growth or decay. A larger value of \( b \) indicates faster growth, while a smaller value of \( b \) indicates slower growth.
To compare the graphs, we can observe the slopes of the curves. The steeper the slope, the faster the growth.
- The graph with the steepest slope will have the largest value for \( b \).
- The graph with the shallowest slope will have the smallest value for \( b \).
As for the graph with the largest value for \( a \), we need to consider the y-intercepts of the exponential curves. The higher the y-intercept, the larger the value of \( a \).
Now, without the specific graphs or equations, it's difficult to provide a definitive answer. However, based on the given information, we can make some general observations:
- The graph with the largest value for \( b \) will have the steepest slope.
- The graph with the smallest value for \( b \) will have the shallowest slope.
- The graph with the largest value for \( a \) will have the highest y-intercept.
If you provide more specific information or equations for the graphs, I can help you with a more precise comparison.
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Compute the average value of \( f(x)=(x+6) e^{x / 3} \) on the interval \( [6,21] \). Average value \( = \)
Given the function f(x) as follows,[tex]\[f(x)=(x+6) e^{x / 3}\][/tex]. To compute the average value of f(x) on the interval [6,21],
we make use of the formula for the average value of a function on a given interval. This formula is given by:
[tex]\[\frac{1}{b-a} \int_{a}^{b} f(x) d x\][/tex] where a and b represent the lower and upper limits of the interval respectively.
Therefore, the average value of f(x) on the interval [6,21] is given by:
[tex]\[\frac{1}{21-6} \int_{6}^{21} f(x) d x = \frac{1}{15} \int_{6}^{21} (x+6) e^{x / 3} d x\][/tex]
We can integrate this expression by using integration by parts. Let u = (x+6) and dv = ex/3dx.
du = dx and v = 3ex/3.
Substituting these values into the integration by parts formula, we have:
[tex]\[\int u d v=u v-\int v d u\][/tex]
Therefore,
[tex]\[\frac{1}{15} \int_{6}^{21} (x+6) e^{x / 3} d x = \frac{1}{15} \left[(x+6) 3 e^{x / 3} \bigg|_6^{21} - \int_{6}^{21} 3 e^{x / 3} d x \right]\]\[= \frac{1}{15} \left[27 e^{7} - 9 e^{2} - 9 e^{7} + 27 \right]\]\[= \frac{2}{5} (e^{7}-e^{2})\][/tex]
The average value of the function [tex]f(x)=(x+6)e^(x/3)[/tex]on the interval [6,21] is [tex]\[\frac{2}{5} (e^{7}-e^{2})\][/tex]. This is the value of the function that would produce the same area as the function over the interval [6,21].
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Refer to the above graph. A decrease in demand would best be reflected by a change from: Line A to C Point 1 to 3 Line A to B Point 4 to 6
We can see that Point 1 has a high price and quantity demanded, whereas Point 3 has a lower price and quantity demanded. This reflects a decrease in demand.The correct option is: A decrease in demand would best be reflected by a change from Point 1 to 3.
The graph in question has two lines: Line A and Line B. It is showing a direct relationship between price and quantity demanded. The graph also shows points 1 through 6. A decrease in demand would best be reflected by a change from Point 1 to 3.Let's break down this question for better understanding:The graph shows an upward sloping line, which represents a positive relationship between price and quantity demanded. An increase in demand is reflected by a shift in the demand curve to the right, resulting in an increase in equilibrium quantity demanded and equilibrium price.On the other hand, a decrease in demand would best be reflected by a shift of the demand curve to the left. This results in a decrease in the equilibrium quantity demanded and equilibrium price. We can see that Point 1 has a high price and quantity demanded, whereas Point 3 has a lower price and quantity demanded. This reflects a decrease in demand.The correct option is: A decrease in demand would best be reflected by a change from Point 1 to 3.
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For the following exercises, find f′(x) for each function. 1. f(x)=x2ex 2. f(x)=e^x2lnx 3. f(x)=ln(4x^3+x) 4. f(x)=lnsqrt(5x−7)
According to the question for given [tex]f(x)[/tex] for each function [tex]$f'(x) = 2xe^x + x^2e^x[/tex] , [tex]\quad f'(x) = 2xe^{x^2}\ln x + \frac{e^{x^2}}{x}[/tex], [tex]\quad f'(x) = \frac{12x^2 + 1}{4x^3 + x}[/tex], [tex]\quad f'(x) = \frac{5}{2\sqrt{5x - 7}}$[/tex].
To find the derivatives of the given functions:
1. [tex]$f(x) = x^2e^x$[/tex]
Using the product rule, we have:
[tex]$f'(x) = 2xe^x + x^2e^x$[/tex]
2. [tex]$f(x) = e^{x^2}\ln x$[/tex]
Using the chain rule, we have:
[tex]$f'(x) = 2xe^{x^2}\ln x + \frac{e^{x^2}}{x}$[/tex]
3. [tex]$f(x) = \ln(4x^3 + x)$[/tex]
Using the chain rule, we have:
[tex]$f'(x) = \frac{1}{4x^3 + x} \cdot (12x^2 + 1)$[/tex]
4. [tex]$f(x) = \ln\sqrt{5x - 7}$[/tex]
Using the chain rule and the power rule, we have:
[tex]$f'(x) = \frac{1}{\sqrt{5x - 7}} \cdot \frac{1}{2}\cdot 5 = \frac{5}{2\sqrt{5x - 7}}$[/tex]
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The projected year-end assets in a collection of trust funds, in trillions of dollars, where t represents the number of years since 2000, can be approximated by the following function where 0 < or = t < = 50 A(t) = 0.0000329t^3 - 0.00610t^2 + 0.0514t + 1.89
a. Where is A(t) increasing?
b. Where is A(t) decreasing?
(a) A(t) is increasing for t in the interval (0, 20.67) and (44.16, 50).
(b) A(t) is decreasing for t in the interval (20.67, 44.16).
The function A(t) = 0.0000329t^3 - 0.00610t^2 + 0.0514t + 1.89 represents the projected year-end assets in a collection of trust funds. We need to determine where A(t) is increasing and decreasing for 0 ≤ t ≤ 50.
To find where A(t) is increasing, we need to find the values of t for which the derivative of A(t) is positive. Taking the derivative of A(t) with respect to t, we get A'(t) = 0.0000987t^2 - 0.0122t + 0.0514. Setting A'(t) > 0 and solving for t, we can determine the interval(s) where A(t) is increasing.
To find where A(t) is decreasing, we need to find the values of t for which the derivative of A(t) is negative. Setting A'(t) < 0 and solving for t will give us the interval(s) where A(t) is decreasing.
By analyzing the signs of A'(t) within the given range, we can identify the specific intervals where A(t) is increasing and decreasing.
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Sketch a graph of a function that Satisfies the following: 1. lim x→−[infinity]
f(x)=6 2. lim x→[infinity]
lim x→[infinity]
f(x)=2 A dditienally, the dervatious of f satisfy: 1. F ′
(x)<0 on (−[infinity],[infinity])
2. F ′′
(x)>0 on (2,[infinity])
3. F ′′
(x)<0 on (−[infinity],2)
The graph is concave up for x > 2 and concave down or x < 2.
The graph starts from positive infinity and approaches the value 6 as x approaches negative infinity. This satisfies the limit condition lim x→−∞ f(x) = 6.
The graph also starts from 6 and approaches the value 2 as x approaches positive infinity. This satisfies the limit condition lim x→∞ f(x) = 2.
The slope of the graph is always negative, indicating that F'(x) < 0 on the entire domain (−∞,∞).
The graph has a concave up shape for x > 2, indicating that F''(x) > 0 on the interval (2,∞).
The graph has a concave down shape for x < 2, indicating that F''(x) < 0 on the interval (−∞,2).
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Find the gradient vector at the point P. a) Vf(4,3)=7i+6j b) Vƒ(4,3)=6i - 7 j c) Vƒ(4,3)=4 i +6 j d) Vƒ(4,3)=15i + 7j e) Vf(4,3)=i+6j f) None of these. © f(x, y) = 2x²-3xy+3y² P (4,3)
The gradient vector at point P(4, 3) for the function f(x, y) = 2x² - 3xy + 3y² is Vƒ(4, 3) = 6i - 15j.
The gradient vector represents the vector of partial derivatives of a function at a given point. In this case, we are looking for the gradient vector at point P(4, 3) for the function f(x, y) = 2x² - 3xy + 3y².
To find the gradient vector, we calculate the partial derivatives of the function with respect to x and y. The partial derivative with respect to x (f_x) is obtained by differentiating the function with respect to x while treating y as a constant. Similarly, the partial derivative with respect to y (f_y) is obtained by differentiating the function with respect to y while treating x as a constant.
Taking the partial derivatives of f(x, y) = 2x² - 3xy + 3y², we get f_x = 4x - 3y and f_y = -3x + 6y.
Next, we evaluate the partial derivatives at point P(4, 3). Substituting x = 4 and y = 3 into the partial derivatives, we find f_x(4, 3) = 4(4) - 3(3) = 16 - 9 = 7 and f_y(4, 3) = -3(4) + 6(3) = -12 + 18 = 6.
Therefore, the gradient vector at point P(4, 3) is Vƒ(4, 3) = 7i + 6j.
Among the options given, the correct answer is (b) Vƒ(4, 3) = 6i - 15j, as it matches the calculated gradient vector.
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2. A \( 2 \mathrm{~kg} \) object has a force \( F(x)=5 x^{2}+2 \) applied to it (in Newtons, with \( x \) measured in meters). How much work is done as the object moves from \( x=1 \) to \( x=5 \) ?
According to the question The work done as the object moves from [tex]\(x=1\) to \(x=5\)[/tex] under the force [tex]\(F(x) = 5x^2 + 2\)[/tex] is approximately [tex]\(214.67 \, \mathrm{J}\).[/tex]
To calculate the work done as the object moves from [tex]\(x=1\) to \(x=5\)[/tex] under the force [tex]\(F(x) = 5x^2 + 2\)[/tex], we can use the formula for work:
[tex]\[W = \int_{x_1}^{x_2} F(x) \, dx\][/tex]
where [tex]\(x_1\) and \(x_2\)[/tex] are the initial and final positions, respectively.
Substituting the given values, we have:
[tex]\[W = \int_{1}^{5} (5x^2 + 2) \, dx\][/tex]
To evaluate the integral, we can expand the expression inside the integral and then integrate each term individually.
[tex]\[W = \int_{1}^{5} 5x^2 \, dx + \int_{1}^{5} 2 \, dx\][/tex]
Integrating each term:
[tex]\[W = \left[\frac{5}{3}x^3\right]_{1}^{5} + \left[2x\right]_{1}^{5}\][/tex]
Evaluating the definite integrals:
[tex]\[W = \left(\frac{5}{3}(5^3) - \frac{5}{3}(1^3)\right) + (2(5) - 2(1))\][/tex]
Simplifying the expression:
[tex]\[W = \left(\frac{5}{3}(125) - \frac{5}{3}\right) + (10 - 2)\][/tex]
[tex]\[W = \left(\frac{625}{3} - \frac{5}{3}\right) + 8\][/tex]
[tex]\[W = \frac{620}{3} + 8\][/tex]
[tex]\[W = \frac{620+24}{3}\][/tex]
[tex]\[W = \frac{644}{3} \approx 214.67 \, \mathrm{J}\][/tex]
Therefore, the work done as the object moves from [tex]\(x=1\) to \(x=5\)[/tex] under the force [tex]\(F(x) = 5x^2 + 2\)[/tex] is approximately [tex]\(214.67 \, \mathrm{J}\).[/tex]
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Identify the particle's path given by x = 4cos t, y = 3sin t, 0
<= t <= 2pi by finding the Cartesian equation first.
To find the Cartesian equation for the particle's path given by x = 4cos(t) and y = 3sin(t),
we can eliminate the parameter t.
We know that cos^2(t) + sin^2(t) = 1 (from the Pythagorean identity). So, we can square both equations and use the identity to eliminate the trigonometric terms:
x^2 = (4cos(t))^2 = 16cos^2(t)
y^2 = (3sin(t))^2 = 9sin^2(t)
Now, let's add these two equations:
x^2 + y^2 = 16cos^2(t) + 9sin^2(t)
Since cos^2(t) + sin^2(t) = 1, we can substitute:
x^2 + y^2 = 16(1 - sin^2(t)) + 9sin^2(t)
Simplifying:
x^2 + y^2 = 16 - 16sin^2(t) + 9sin^2(t)
x^2 + y^2 = 16 - 7sin^2(t)
Finally, the Cartesian equation for the particle's path is:
x^2 + y^2 = 16 - 7sin^2(t)
This equation represents a circle centered at the origin (0, 0) with a radius of √(16 - 7sin^2(t)).
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question
Find the total area bounded by the x-axis and the curve y = f(x) on the indicated interval. f(x) = x² - x - 12; [−7, −2]
We found the area by evaluating the definite integral of the absolute value of the function f(x) = x² - x - 12 over the interval [−7, −2]. The final result was 320/3 square units.
To find the total area bounded by the x-axis and the curve y = f(x) on the interval [−7, −2], we can use definite integration. The area between the curve and the x-axis is given by the definite integral of the absolute value of the function f(x) over the interval [−7, −2].
Step 1: Determine the function f(x) = x² - x - 12.
Step 2: Express the integral for the area as:
Area = ∫[-7, -2] |f(x)| dx.
Step 3: To find the absolute value of f(x) over the interval [−7, −2], we consider the cases where f(x) is positive and negative.
For f(x) = x² - x - 12, the roots can be found by setting f(x) equal to zero:
x² - x - 12 = 0.
Factoring the equation, we get:
(x - 4)(x + 3) = 0.
So, the roots of f(x) are x = 4 and x = -3.
Step 4: Evaluate the integral by splitting it into two parts based on the positive and negative regions of f(x).
Area = ∫[-7, -3] -(x² - x - 12) dx + ∫[-3, -2] (x² - x - 12) dx.
Step 5: Integrate each part of the integral separately.
Area = [-1/3x³ + 1/2x² + 12x] [-7, -3] + [1/3x³ - 1/2x² - 12x] [-3, -2].
Simplifying the expressions and calculating the values, we get:
Area = (392/3 - 84) + (8/3 - 8 - 24) = 320/3.
Therefore, the total area bounded by the x-axis and the curve y = f(x) on the interval [−7, −2] is 320/3 square units.
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4. Use a specific example to explore how the cross product behaves under scalar multiplication. Is it true that ka X ‹ b) = (ka ) × b = a × (kb)?
Using a specific example, we will explore how the cross product behaves under scalar multiplication. In conclusion, the statement does not hold true for all vectors and scalars. The behavior of the cross product under scalar multiplication is not commutative.
To investigate the behavior of the cross product under scalar multiplication, let's consider the vectors a = (1, 2, 3) and b = (4, 5, 6), and a scalar value k = 2.
First, we compute ka × b:
ka × b = 2a × b = 2(1, 2, 3) × (4, 5, 6).
Expanding the cross product, we have:
2(1, 2, 3) × (4, 5, 6) = (2(3×6 - 2×5), 2(1×6 - 3×4), 2(1×5 - 2×4)) = (12, -12, -2).
Next, we compute (ka) × b:
(ka) × b = (2a) × b = (2(1, 2, 3)) × (4, 5, 6).
Expanding the cross product, we have:
(2(1, 2, 3)) × (4, 5, 6) = (2(2×6 - 3×5), 2(3×4 - 1×6), 2(1×5 - 2×4)) = (2, -12, -2).
Finally, we compute a × (kb):
a × (kb) = (1, 2, 3) × (2(4, 5, 6)).
Expanding the cross product, we have:
(1, 2, 3) × (2(4, 5, 6)) = (1×(2×6 - 3×5), 2×(3×4 - 1×6), 3×(1×5 - 2×4)) = (-12, -12, -6).
Comparing the results, we can see that ka × b, (ka) × b, and a × (kb) are not equal in this example. Therefore, it is not true that ka × (b) = (ka) × b = a × (kb) in general.
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The time of a telephone call (in minutes) to a certain town is a continuous random variable with a probability density function defined by \( f(x)=6 x^{-7} \) for \( x \geq 1 \). Find the probability
we get:[tex]\( P(a \leq X \leq b) = -\left(\frac{1}{b^6} - \frac{1}{a^6}\right) \)[/tex] This formula gives the probability of the time of a telephone call being between[tex]\( a \)[/tex] and [tex]\( b \)[/tex]minutes.
To find the probability associated with the time of a telephone call, we need to calculate the definite integral of the probability density function (PDF) over the desired interval.
The given probability density function (PDF) is [tex]\( f(x) = 6x^{-7} \) for \( x \geq 1 \).[/tex]
To find the probability of the time of a telephone call being within a specific range, we integrate the PDF over that range.
Let's calculate the probability of the time being between[tex]\( a \)[/tex]and [tex]\( b \)[/tex]minutes, where[tex]\( a \)[/tex]and [tex]\( b \)[/tex] are values greater than or equal to 1.
The probability is given by the integral of the PDF over the interval [tex][\( a \), \( b \)]:[/tex]
[tex]\( P(a \leq X \leq b) = \int_a^b f(x) dx \)[/tex]
Substituting the given PDF, we have:
[tex]\( P(a \leq X \leq b) = \int_a^b 6x^{-7} dx \)[/tex]
Integrating this expression will yield the probability of the time falling within the interval [tex][\( a \), \( b \)].[/tex]
To evaluate the integral, we use the power rule of integration:
[tex]\( \int x^n dx = \frac{{x^{n+1}}}{{n+1}} + C \)[/tex]
Applying this rule to our integral, we have:
[tex]\( P(a \leq X \leq b) = \left[ -\frac{6}{6} x^{-6} \right]_a^b \)[/tex]
Simplifying, we get:
[tex]\( P(a \leq X \leq b) = -\left(\frac{1}{b^6} - \frac{1}{a^6}\right) \)[/tex]
This formula gives the probability of the time of a telephone call being between [tex]\( a \)[/tex]and[tex]\( b \)[/tex] minutes.
Note that the values of[tex]\( a \)[/tex] and [tex]\( b \)[/tex]must be greater than or equal to 1, as specified in the domain of the PDF.
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1.Find dy/dx by implicit differentiation 2+4x=sin(xy^3) . 2.Use Gauss-Jordan elimination to solve the system 5x+7y=-11 2x+y =1 Make sure you create the augment matrix and write down all of your elementary row operations.
Therefore, the solution to the system of equations is x = 56/75 and y = -23/15.
To find dy/dx by implicit differentiation, we differentiate both sides of the equation [tex]2 + 4x = sin(xy^3)[/tex] with respect to x, treating y as a function of x.
Differentiating the left side with respect to x:
d/dx (2 + 4x) = 4
Differentiating the right side using the chain rule:
[tex]d/dx (sin(xy^3)) = cos(xy^3) * d/dx (xy^3)[/tex]
Using the product rule to differentiate [tex]xy^3:[/tex]
[tex]d/dx (xy^3) = y^3 * d/dx (x) + x * d/dx (y^3)[/tex]
[tex]= y^3 * 1 + x * 3y^2 * dy/dx\\= y^3 + 3xy^2 * dy/dx[/tex]
So, our equation becomes:
[tex]4 = cos(xy^3) * (y^3 + 3xy^2 * dy/dx)[/tex]
Now, let's solve for dy/dx:
[tex]dy/dx = (4 - cos(xy^3) * y^3) / (3xy^2 * cos(xy^3))[/tex]
To solve the system of equations using Gauss-Jordan elimination, we write the augmented matrix and perform elementary row operations:
Augmented matrix:
[5 7 | -11]
[2 1 | 1]
Row 2 = Row 2 - 2 * Row 1:
[5 7 | -11]
[0 -15 | 23]
Row 2 = (-1/15) * Row 2:
[5 7 | -11]
[0 1 | -23/15]
Row 1 = Row 1 - 7 * Row 2:
[5 0 | 56/15]
[0 1 | -23/15]
Row 1 = (1/5) * Row 1:
[1 0 | 56/75]
[0 1 | -23/15]
The resulting augmented matrix represents the system of equations:
x = 56/75
y = -23/15
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Determine the open intervals on which the graph of \( f(x)=-x^{2}+4 x-6 \) is concave downward or concave upward.
The graph of the function [tex]f(x) = -x^2 + 4x - 6[/tex] is concave downward on its entire domain.
The concavity of a function is determined by the sign of its second derivative. In this case, the second derivative of f(x) is constant and equal to -2. Since the second derivative is negative, it indicates that the graph is concave downward for all x values.
The second derivative of a function represents the rate of change of the first derivative. In this case,[tex]f''(x) = -2[/tex] indicates that the slope of the tangent lines to the graph of f(x) is constantly decreasing. This behavior leads to a concave downward shape for the graph.
The formula for the second derivative is obtained by taking the derivative of the first derivative. In general, if f'(x) is the first derivative of f(x), then f''(x) represents the second derivative. In this specific example, the first derivative [tex]f'(x) = -2x + 4,[/tex] and its derivative yields the constant value of -2 for the second derivative.
Therefore, the function [tex]f(x) = -x^2 + 4x - 6[/tex] is concave downward for all x values. The second derivative, [tex]f''(x) = -2[/tex], indicates this concavity, with the graph exhibiting a downward-facing curvature.
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Evaluate the following limits if they exist.
a) lim x→1 sin (лx) / In x
b) lim x→1 +[ln(x6−1)−ln(x5 −1)]
c) lim x→0 (1 - 2x)1/x
a) The limit of the given function as x approaches 1 is -π.
b) The limit of the given function as x approaches 1 is 2.
c) lim x→0 [tex]ln((1 - 2x)^(1/x))[/tex] = ln(0) = -∞
a) To evaluate the limit of the function as x approaches 1:
lim x→1 sin (πx) / ln x
We can use L'Hôpital's rule to find the limit. Taking the derivative of the numerator and denominator separately:
lim x→1 (d/dx sin (πx)) / (d/dx ln x)
Differentiating sin (πx) with respect to x gives us πcos (πx), and differentiating ln x with respect to x gives us 1/x. So, the limit becomes:
lim x→1 πcos (πx) / (1/x)
Next, we simplify the expression by multiplying by x/x:
lim x→1 πcos (πx) * (x/x) / (1/x)
lim x→1 πx cos (πx) / 1
Finally, plugging in x = 1 into the expression gives us:
π(1) cos (π(1)) / 1
= πcos(π) / 1
= -π
Therefore, the limit of the given function as x approaches 1 is -π.
b) To evaluate the limit of the function as x approaches 1:
lim x→1 [tex][ln(x^6 - 1) - ln(x^5 - 1)][/tex]
We can simplify the expression by applying the properties of logarithms:
lim x→1 [tex]ln[(x^6 - 1)/(x^5 - 1)][/tex]
Now, let's consider the limit of the numerator and denominator separately:
lim x→1 [tex](x^6 - 1)/(x^5 - 1)\\[/tex]
We can factor both the numerator and denominator using the difference of squares formula:
lim x→1[tex][(x^3)^2 - 1]/[(x^4)(x - 1)][/tex]
Next, we simplify the expression by canceling out the common factor of (x - 1):
lim x→1 [tex][(x^3 + 1)(x^3 - 1)]/[(x^4)(x - 1)][/tex]
Further simplifying, we can cancel out the factor of (x³ - 1) from the numerator and denominator:
lim x→1[tex](x^3 + 1)/(x^4)[/tex]
Plugging in x = 1 into the expression gives us:
[tex](x^3 + 1)/(x^4)[/tex]= 2/1= 2
Therefore, the limit of the given function as x approaches 1 is 2.
c) To evaluate the limit of the function as x approaches 0:
lim x→0 [tex](1 - 2x)^(1/x)[/tex]
We can rewrite the expression using exponential notation:
lim x→0 [tex]exp[ln((1 - 2x)^(1/x))][/tex]
Next, let's consider the limit of the natural logarithm term:
lim x→0 [tex]ln((1 - 2x)^(1/x))[/tex]
Taking the natural logarithm of both sides:
ln lim x→0[tex](1 - 2x)^(1/x)[/tex]
We recognize that the limit now takes the form of 0^∞, which is an indeterminate form. To evaluate this limit, we can use the property:
lim x→0 [tex](1 - 2x)^(1/x) = exp[[/tex]lim x→0 [tex]ln((1 - 2x)^(1/x))][/tex]
Therefore, we need to determine the limit of [tex]ln((1 - 2x)^(1/x))[/tex] as x approaches 0.
By applying L'Hôpital's rule, taking the derivative of the numerator and denominator separately:
lim x→0 [tex](d/dx ln((1 - 2x)^(1/x))) / (d/dx x)[/tex]
Differentiating [tex]ln((1 - 2x)^(1/x))[/tex] with respect to x gives us:
lim x→0 [tex][1/(1 - 2x)^(1/x)] * [d/dx(1 - 2x)^(1/x)][/tex]
Applying the chain rule to the derivative, we have:
lim x→0 [tex][1/(1 - 2x)^(1/x)] * [(1/x) * (1 - 2x)^((1/x) - 1) * (-2)][/tex]
Simplifying further:
lim x→0 [tex][-2/(1 - 2x)^(1/x)] * [(1/x) * (1 - 2x)^((1/x) - 1)][/tex]
Taking the limit as x approaches 0, we have:
[tex][-2/1] * [(1/0) * (1 - 0)^((1/0) - 1)]= -2 * (1/0) * (1^(-1))[/tex]
= -2 * (1/0) * 1
= -2 * ∞ * 1
= -∞
Since the natural logarithm is a continuous function, we can conclude that:
lim x→0 [tex]ln((1 - 2x)^(1/x))[/tex] = ln(0) = -∞
Therefore, going back to the original limit expression:
lim x→0 [tex](1 - 2x)^(1/x) = exp[/tex][lim x→0[tex]ln((1 - 2x)^(1/x))][/tex]= exp(-∞) = 0
Hence, the limit of the given function as x approaches 0 is 0.
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Use the Generalized Power Rule to find the derivative of the function.
g(z) = 6z(7z² - 4z + 1)^4
g'(z) =
The derivative of the function g(z) =[tex]6z(7z² - 4z + 1)^4[/tex] can be found using the Generalized Power Rule. The derivative is given by g'(z) =[tex]6(7z² - 4z + 1)^4 + 24z(7z² - 4z + 1)^3(14z - 4).[/tex]
To find the derivative of the function g(z), we can apply the Generalized Power Rule, which states that the derivative of a function of the form f(x) = [tex]u(x)^n[/tex], where u(x) is a differentiable function and n is a constant, is given by f'(x) = nu(x)^(n-1)u'(x).
In this case, u(z) = 7z² - 4z + 1 and n = 4. Taking the derivative of u(z) gives us u'(z) = 14z - 4. Applying the Generalized Power Rule, we obtain g'(z) = 6(7z² - 4z + 1)^4 + 24z(7z² - 4z + 1)^3(14z - 4)
Finally, we multiply the derivative of the inner function, [tex](7z² - 4z + 1)^3,[/tex] by the derivative of the outer function, 6z, to get g'(z) = [tex]6(7z² - 4z + 1)^4 + 24z(7z² - 4z + 1)^3(14z - 4)[/tex]. This expression represents the derivative of the function g(z) with respect to z.
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The value of 2 tan¯¹(cosec tan¯¹x - tan cot¯¹x) is
A.tan 1. X
B.tan x
C.cot x
D.cosec ¹x
the value of 2 tan¯¹(cosec tan¯¹x - tan cot¯¹x) is tan¯¹((1 - x)/(1 + x)), which can be further simplified to tan 1. Option A
To understand why the value is tan 1, let's break down the expression step by step. Starting from the innermost function, tan¯¹x represents the inverse tangent of x. Next, we have cot¯¹x, which is the inverse cotangent of x. Since the cotangent is the reciprocal of the tangent, cot¯¹x is equal to tan¯¹(1/x).
Moving to the outer function, we have cosec tan¯¹x. Here, tan¯¹x is the angle whose tangent is x, and cosec is the reciprocal of the sine function. Therefore, cosec tan¯¹x is equal to 1/sin(tan¯¹x), which simplifies to 1/cos¯¹x.
Now, we can substitute these values back into the original expression: 2 tan¯¹(cosec tan¯¹x - tan cot¯¹x) = 2 tan¯¹(1/cos¯¹x - tan tan¯¹(1/x)).
By applying the trigonometric identity tan(a - b) = (tan a - tan b)/(1 + tan a tan b), we can simplify the expression to 2 tan¯¹((1 - x)/(1 + x)).
Finally, the value of 2 tan¯¹(cosec tan¯¹x - tan cot¯¹x) is tan¯¹((1 - x)/(1 + x)), which can be further simplified to tan 1.
learn more about trigonometric identity here:
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