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) = (1 − x)³

Accepted Answer

First, recall that to show a function has an inverse over its domain, it must be one-to-one (injective) and continuous on that domain. For the function $$f(x) = (1 - x)^3$, observe that it is a cubic function shifted and reflected, which is strictly monotonic (either strictly increasing or decreasing) over all real numbers, so it is one-to-one over its entire domain. Next, find the derivative of f(x$$)$$ with respect to x$. Using the chain rule, we have f$$'(x) = 3(1 - $$x)^2$$ \cdot (-1) = -3(1 - $$x)^2. $$According to Theorem 1 (the formula for the derivative of the inverse function), if $$y = f(x)$$ has an inverse $$f^{-1}$$, then the derivative of the inverse function at $$y is $$given by $$\frac{d}{dy} f$$^{-1}$$(y) = \frac{1}{f'(f$$^{-1}$$(y))}. To $$find a formula for $$\frac{d}{dx} f$$^{-1}$$(x)$$, replace $$y by$$ $$x in $$the formula above, so $$\frac{d}{dx} f$$^{-1}$$(x) = \frac{1}{f'(f$$^{-1}$$(x))}. $$This means you need to express the derivative of $$f$$ evaluated at $$f^{-1}(x$$)$$. Finally, substitute the expression for f$$'(x)$$ into the formula: $$\frac{d}{dx} $$f^{-1}(x$$) = \frac{1}{-3(1 - $$f^{-1}(x))^2$}. $$This gives the derivative of the inverse function in terms of $$f^{-1}(x$$)$.

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) = (1 − x)³

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Key Concepts

Invertibility of Functions

Derivative of the Inverse Function

Theorem 1 (Inverse Function Theorem)