Ch. 7 - Transcendental Functions

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 7 - Transcendental Functions

Problem 7.1.72d

Chapter 7, Problem 7.1.72d

In Exercises 67–72, you will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS:
d. Find the equation for the tangent line to g at the point (f(x_0), x_0) located symmetrically across the 45° line y=x (which is the graph of the identity function). Use Theorem 1 to find the slope of this tangent line.
72. y= 2-x-x³, -2 ≤ x ≤ 2, x_0 = 3/2

Verified step by step guidance

Identify the given function as \( y = f(x) = 2 - x - x^{3} \) with the domain \( -2 \leq x \leq 2 \) and the point \( x_0 = \frac{3}{2} \).

Calculate \( f(x_0) \) by substituting \( x_0 \) into the function: \( f\left( \frac{3}{2} \right) = 2 - \frac{3}{2} - \left( \frac{3}{2} \right)^3 \). This gives the \( y \)-coordinate of the point on \( f \).

Recall that the inverse function \( g = f^{-1} \) swaps the roles of \( x \) and \( y \), so the point on \( g \) corresponding to \( x_0 \) is \( (f(x_0), x_0) \).

Use Theorem 1, which states that if \( g = f^{-1} \), then the derivative of \( g \) at \( y = f(x_0) \) is \( g'(f(x_0)) = \frac{1}{f'(x_0)} \). First, find \( f'(x) = \frac{d}{dx} (2 - x - x^3) = -1 - 3x^2 \), then evaluate \( f'(x_0) \).

Write the equation of the tangent line to \( g \) at the point \( (f(x_0), x_0) \) using the point-slope form: \[ y - x_0 = g'(f(x_0)) (x - f(x_0)) \]. Substitute the values found for \( x_0 \), \( f(x_0) \), and \( g'(f(x_0)) \) to express the tangent line equation.

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Key Concepts

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

Inverse Functions and Their Graphs

An inverse function reverses the roles of inputs and outputs of the original function, swapping x and y values. Graphically, the inverse function is the reflection of the original function across the line y = x (the 45° line). Understanding this symmetry is crucial for locating points and tangent lines on the inverse function.

Derivative of an Inverse Function (Theorem 1)

Theorem 1 states that if a function f is differentiable and invertible at x₀, then the derivative of its inverse g at y₀ = f(x₀) is given by g'(y₀) = 1 / f'(x₀). This relationship allows us to find the slope of the tangent line to the inverse function using the derivative of the original function.

Equation of a Tangent Line

The tangent line to a function at a point provides the best linear approximation near that point. Its equation is y - y₁ = m(x - x₁), where m is the slope (derivative at the point) and (x₁, y₁) is the point of tangency. For inverse functions, the point and slope must be carefully identified using the symmetry and derivative relationships.

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In Exercises 67–72, you will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS:

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