Question

Use the results of Exercise 55 to show that the functions in Exercises 56–60 have inverses over their domains. Find a formula for df⁻¹/dx using Theorem 1.f(x) = 27x³

Accepted Answer

First, recall that a function has an inverse if it is one-to-one (injective) over its domain. Since the function is $$f(x) = 27x^3$, note that the cubic function x^3$ is strictly increasing over all real numbers, so f(x$$)$$ is also strictly increasing and therefore one-to-one over its domain. Next, to find the derivative of the inverse function f$$^{-1}$$(x)$$, use Theorem 1, which states that if $$f is $$differentiable and has an inverse, then the derivative of the inverse at a point $$x is $$given by:

$$\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}$$ Calculate the derivative of the original function $$f(x)$$:

$$f'(x) = \frac{d}{dx}(27x^3) = 81x^2$$ Express the formula for the derivative of the inverse function by substituting $$f'(x)$$ into the formula from Theorem 1:

$$\frac{d}{dx}f^{-1}(x) = \frac{1}{81 (f^{-1}(x))^2}$$ Finally, to write the derivative of the inverse explicitly in terms of $$x$$, find the inverse function $$f^{-1}(x$$)$$ by solving y$$ = $$27x^3$$ for $$x$$:

$$x = \sqrt[3]{\frac{y}{27}} = \frac{\sqrt[3]{y}}{3}$$

Replace $$y$$ with $$x to $$get $$f^{-1}(x$$) = \frac{\sqrt[3]{x}}{3}$$, and substitute this into the derivative formula to express $$\frac{d}{dx}$$f^{-1}(x$$)$$ fully in terms of x$.

Use the results of Exercise 55 to show that the functions in Exercises 56–60 have inverses over their domains. Find a formula for df⁻¹/dx using Theorem 1.
f(x) = 27x³

Verified video answer for a similar problem:

Key Concepts

Inverse Functions

One-to-One and Monotonicity

Derivative of the Inverse Function (Theorem 1)