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.

70. y= x³/(x²+1), -1 ≤ x ≤ 1, x_0=1/2

Accepted Answer

Identify the given function as $$ y = f(x) = \frac{x^3}{x^2 + 1} $$ and the point $$ x_0 = \frac{1}{2} . $$The function $$ g  is $$the inverse of $$ f $$, so $$ g = f^{-1} . $$The point on $$ g $$ corresponding to $$ x_0  is$$ $$ (f(x_0), x_0) . $$Calculate $$ f(x_0)  by $$substituting $$ x_0 = \frac{1}{2} $$ into $$ f(x) $$:

$$
f\left(\frac{1}{2}\right) = \frac{\left(\frac{1}{2}\right)^3}{\left(\frac{1}{2}\right)^2 + 1}
$$ Find the derivative $$ f'(x) $$ using the quotient rule:

$$
f'(x) = \frac{(3x^2)(x^2 + 1) - x^3(2x)}{(x^2 + 1)^2}
$$
Then evaluate $$ f'(x_0)  at$$ $$ x_0 = \frac{1}{2} . $$Use Theorem 1, which states that the derivative of the inverse function at $$ y_0 = f(x_0)  is $$the reciprocal of the derivative of the original function at $$ x_0 $$:

$$
g'(y_0) = \frac{1}{f'(x_0)}
$$
This gives the slope of the tangent line to $$ g  at $$the point $$ (f(x_0), x_0) . $$Write the equation of the tangent line to $$ g  at$$ $$ (f(x_0), x_0) $$ using point-slope form:

$$
 y - x_0 = g'(f(x_0)) (x - f(x_0))
$$
where $$ y $$ and $$ x $$ are the coordinates on the graph of $$ g .$$

Verified video answer for a similar problem:

Key Concepts

Inverse Functions and Their Graphs

Derivative of an Inverse Function (Theorem 1)

Tangent Line Equation and Point-Slope Form