For the function $f (x) = {(x - 3)}^{3} + 1$ $f(x)=(x-3)^3+1$ , how do you find $f^{- 1} (x)$ $f^-1(x)$ ?

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For the function $f (x) = {(x - 3)}^{3} + 1$ $f(x)=(x-3)^3+1$ , how do you find $f^{- 1} (x)$ $f^-1(x)$ ?

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Answer 1 · 2017-02-28T21:37:06

In order to find for any function $f (x)$ $f(x)$ , we must apply the 'transformation' $y = x$ $y=x$ . In order to do that, we must define $x$ $x$ in terms of $y$ $y$ , ie., find $f (y)$ $f(y)$ , then set $y = x$ $y=x$ .

Let $y = f (x)$ $y=f(x)$

$y = {(x - 3)}^{3} + 1$ $y=(x-3)^3+1$

$y - 1 = {(x - 3)}^{3}$ $y-1=(x-3)^3$

$x - 3 = \sqrt[3]{y - 1}$ $x-3=root(3)(y-1)$

$x = 3 + \sqrt[3]{y - 1}$ $x=3+root(3)(y-1)$

We've now found $f (y)$ $f(y)$ , so we must set $y = x$ $y=x$ by replacing $x$ $x$ with $y$ $y$ and $y$ $y$ with $x$ $x$ .

$f^{- 1} (x) = y = 3 + \sqrt[3]{x - 1}$ $f^-1(x)=y=3+root(3)(x-1)$

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For the function $f (x) = {(x - 3)}^{3} + 1$ $f(x)=(x-3)^3+1$ , how do you find $f^{- 1} (x)$ $f^-1(x)$ ?

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