Thread: Showing one to one

f(x) = 2x^3 - (cosx)^3, x for all [0, pi]. Show f is one to one and find (f^-1)'(-1). Note (0,-1) for all f.

x = 2y^3 - (cosy)^3

Is there a way I can factor the y so there is only one of them?

On $\displaystyle \left( {0,\pi } \right)$ observe that $\displaystyle f'(x) \ge 0$. That means that $\displaystyle f$ is increasing there.
Moreover, $\displaystyle f(0) = - 1$.