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

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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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주요 개념

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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.