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

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The discussion focuses on proving that the function f(x) = x^3 - 1 is one-to-one by demonstrating that if f(s) = f(t), then s must equal t, confirming injectivity. It also establishes that f is surjective, as for any real number y, there exists an x such that y = f(x). Consequently, f is bijective, allowing the definition of its inverse function, f^{-1}(x) = √[3]{x^3 + 1}. The derivative of the inverse function is derived as (f^{-1})'(x) = x^2 / √[3]{(x^3 + 1)^2} for x ≠ -1. This comprehensive explanation aids in understanding the properties of the function and its inverse.
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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, I am 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.
 
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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, $$\left(f^{-1}\right)'(x)=\ldots=\frac{x^2}{\sqrt[3]{(x^3+1)^2}}\qquad (x\neq -1)$$
 
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