For the function f(x)=(x-3)^3+1 find f^-1(x) ?

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Answer 1 · 2018-01-08T08:24:58

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

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

or

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

To find the inverse function $f^{- 1} (x)$ $f^-1(x)$ , we arrange the equation in terms of $x$ $x$ .

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

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

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

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

So, the inverse of $f (x) = {(x - 3)}^{3} + 1$ $f(x)=(x-3)^3+1$ is

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