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

Verified step by step guidance

1. 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 \( (f(x_0), x_0) \).
2. Calculate \( f(x_0) \) by substituting \( x_0 = \frac{1}{2} \) into \( f(x) \): \[ f\left(\frac{1}{2}\right) = \frac{\left(\frac{1}{2}\right)^3}{\left(\frac{1}{2}\right)^2 + 1} \]
3. 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 \( x_0 = \frac{1}{2} \).
4. 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 \( (f(x_0), x_0) \).
5. 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 \( g \).

Key Concepts

Inverse Functions and Their Graphs

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

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.

Tangent Line Equation and Point-Slope Form

The tangent line to a curve at a point is the best linear approximation of the function near that point. Its equation can be found using the point-slope form: y - y₁ = m(x - x₁), where m is the slope (derivative at the point) and (x₁, y₁) is the point of tangency. This is essential for writing the tangent line to g at (f(x₀), x₀).