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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 3 - DerivativesProblem 3.10.86c
Chapter 3, Problem 3.10.86c

Tangents and inverses Suppose L(x)=ax+b (with a≠0) is the equation of the line tangent to the graph of a one-to-one function f at (x0,y0). Also, suppose M(x)=cx+d is the equation of the line tangent to the graph of f^−1 at (y0,x0).


c. Prove that L^−1(x)=M(x).

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1
Start by understanding the relationship between a function and its inverse. If L(x) is tangent to f at (x0, y0), then the slope of L(x), which is 'a', is equal to the derivative of f at x0, i.e., f'(x0) = a.
For the inverse function f^−1, the derivative at a point (y0, x0) is the reciprocal of the derivative of f at x0. Therefore, the slope of M(x), which is 'c', is equal to 1/a, i.e., c = 1/a.
The line L(x) = ax + b is tangent to f at (x0, y0), so it passes through the point (x0, y0). Therefore, y0 = ax0 + b, which can be rearranged to find b: b = y0 - ax0.
Similarly, the line M(x) = cx + d is tangent to f^−1 at (y0, x0), so it passes through the point (y0, x0). Therefore, x0 = cy0 + d, which can be rearranged to find d: d = x0 - cy0.
To prove L^−1(x) = M(x), find the inverse of L(x). The inverse L^−1(x) is found by swapping x and y in the equation y = ax + b and solving for y. This gives L^−1(x) = (x - b)/a. Substitute b = y0 - ax0 into this equation to show that L^−1(x) = M(x) = (1/a)x + (x0 - (1/a)y0).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Inverse Functions

An inverse function essentially reverses the effect of the original function. If f(x) takes an input x and produces an output y, then the inverse function f^−1(y) takes y back to x. For a function to have an inverse, it must be one-to-one, meaning each output is produced by exactly one input.
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Inverse Cosine

Tangent Lines

A tangent line to a function at a given point represents the instantaneous rate of change of the function at that point. The slope of the tangent line is given by the derivative of the function at that point. In the context of the problem, the tangent lines L(x) and M(x) represent the slopes of the original function f and its inverse f^−1, respectively.
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Properties of Derivatives

The derivative of a function at a point provides information about the function's behavior near that point, including its slope. For inverse functions, a key property is that the slopes of the tangent lines at corresponding points are reciprocals of each other. This relationship is crucial for proving that the inverse of the tangent line L(x) equals the tangent line M(x).
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Related Practice
Textbook Question

97–100. Logistic growth Scientists often use the logistic growth function P(t) = P₀K / P₀+(K−P₀)e^−r₀t to model population growth, where P₀ is the initial population at time t=0, K is the carrying capacity, and r₀ is the base growth rate. The carrying capacity is a theoretical upper bound on the total population that the surrounding environment can support. The figure shows the sigmoid (S-shaped) curve associated with a typical logistic model. <IMAGE>


{Use of Tech} Gone fishing When a reservoir is created by a new dam, 50 fish are introduced into the reservoir, which has an estimated carrying capacity of 8000 fish. A logistic model of the fish population is P(t) = 400,000 / 50+7950e^−0.5t, where t is measured in years.


d. Graph P' and use the graph to estimate the year in which the population is growing fastest. 

Textbook Question

62–65. {Use of Tech} Graphing f and f'

c. Verify that the zeros of f' correspond to points at which f has a horizontal tangent line.

f(x)=(x²−1)sin^−1 x on [−1,1]

Textbook Question

Airline travel The following figure shows the position function of an airliner on an out-and-back trip from Seattle to Minneapolis, where s = f(t) is the number of ground miles from Seattle t hours after take-off at 6:00 A.M. The plane returns to Seattle 8.5 hours later at 2:30 P.M. <IMAGE>

d. Determine the velocity of the airliner at noon (t = 6) and explain why the velocity is negative.

Textbook Question

Derivatives using tables Let h(x)=f(g(x))h(x)=f(g(x)) and p(x)=g(f(x))p(x)=g(f(x)). Use the table to compute the following derivatives.

<IMAGE>

d. p′(2)p^{\(\prime\)}\(\left\)(2\(\right\))

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Textbook Question

Vibrations of a spring Suppose an object of mass m is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position y=0y=0 when the mass hangs at rest. Suppose you push the mass to a position y0y_0 units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system), the position y of the mass after t seconds is y=y0cos(tkm)y=y_0\(\cos\)\(\left\)(t\(\sqrt{\frac{k}{m}\)}\(\right\)), where k>0k>0 is a constant measuring the stiffness of the spring (the larger the value of kk, the stiffer the spring) and yy is positive in the upward direction.

Use equation (4) to answer the following questions.

c. How would the velocity be affected if the experiment were repeated with a spring having four times the stiffness (kk is increased by a factor of 44)?

Textbook Question

Finding derivatives from a table Find the values of the following derivatives using the table. <IMAGE>


c. d/dx ((f(x)g(x)) |x=3

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