$$f(x)=x^3+1$$ To find inverse, from what I've learned we change the y to x
$$x=y^3+1$$ solve for y $$x-1=y^3$$
Should I cube root the x-1 for this? if i did that I still would not get the answer that would match the answer choice given to me
What should i do after this?
$\endgroup$ 24 Answers
$\begingroup$$$f(x)=x^3+1$$$$y=x^3+1$$$$y-1=x^3$$$$x=(y-1)^{\frac13}$$$$f^{-1}(x)=(x-1)^{\frac13}$$
$\endgroup$ $\begingroup$$(x-1)^{1/3}$ or $(x-1)^{1/3} \omega$ or $(x-1)^{1/3} \omega^2$, where $\omega = \frac{-1+i\sqrt{3}}{2}$.
The function is invertible over reals - and the inverse is $(x-1)^{1/3}$, however, the function is not invertible in the complex domain since there is no unique inverse.
$\endgroup$ $\begingroup$Inverting the graph of $f$ (red curve) means mirroring at the line $\DeclareMathOperator{id}{id}\id(x)=x$ (blue line).
It can often happen that the resulting graph is not the graph of a function, because it turns multiple valued at some $x$-coordinates. But here (green curve) everything is fine.
This shows up in your correct algebraic invertation, where the last step would be $$ y = \sqrt[3]{x - 1} = (x - 1)^{1/3} $$
$\endgroup$ $\begingroup$$f(x)=x^3+1$
$x^3+1=y$
$x=\sqrt[3]{y-1}$
$\endgroup$
