Question

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

Accepted Answer

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

Inverse Functions and Their Graphs

Derivative of an Inverse Function (Theorem 1)

Equation of a Tangent Line