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Tia's question at Yahoo! Answers (Derivative of f^{-1})

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Here is the question:

**Show that f is one-to-one on its domain.

**Find the derivative of f^-1, where f^-1 is the inverse function of f.

The "-1" is not being raised to anything in the first part (meaning f(x)=x^3-1) by the way, just the "3", please help with this and show how to work it, Im soo lost, would really appreciate it. Thanks.
Here is a link to the question:

Consider the function f(x)=x^3-1...? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello Tia, $$\begin{aligned}f(s)=f(t)&\Rightarrow s^3-1=t^3-1\\&\Rightarrow s^3=t^3\\&\Rightarrow \sqrt[3]{s^3}=\sqrt[3]{t^3}\\&\Rightarrow s=t\\&\Rightarrow f\mbox{ is injective}\end{aligned}$$ On the other hand, consider $y\in\mathbb{R}$. Let's see that there exists $x\in\mathbb{R}$ such that $y=f(x)$ (i.e. $f$ is surjective) $$\begin{aligned}y=f(x)&\Leftrightarrow y=x^3-1\\&\Leftrightarrow x^3=y+1\\&\Leftrightarrow x=\sqrt[3]{y^3+1}\end{aligned}$$ So, $f:\mathbb{R}\to \mathbb{R}$ is bijective and $f^{-1}(x)=\sqrt[3]{x^3+1}$. Then, $$\left(f^{-1}\right)'(x)=\ldots=\frac{x^2}{\sqrt[3]{(x^3+1)^2}}\qquad (x\neq -1)$$