How do you find the inverse of $f(x)=x^3+1$ and graph both f and $f^-1$ ?

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Answer 1 · 2017-01-23T15:27:28

THe answer is $root(3)(x-1)$

Let $y=x^3+1$

$x^3=y-1$

$x=root(3)(y-1)$

Therefore,

$f^-1(x)=root(3)(x-1)$

Verification :

$f(f^-1(x))=f(root(3)(x-1))$

$=(root(3)(x-1))^3+1$

$=x-1+1=x$

The graphs of $f(x)$ and $f^-1(x)$ are symmetric wrt $y=x$

graph{(y-x^3-1)(y-root(3)(x-1))(y-x)=0 [-6.243, 6.243, -3.12, 3.123]}

How do you find the inverse of f(x)=x^3+1f(x)=x^3+1 and graph both f and f^-1f^-1?