The derivative of $f$ is
$$f'(x)=3x^2-18x+33=3(x^2-6x+11)=3[(x-3)^2+2]$$
which is positive. Therefore, $f$ is strictly increasing and hence injective. Nevertheless, an algebraic expression for the inverse it is complicated, because it involves solving of a cubic equation.
7
Bumbble Comm
On
Very simple assume that $f(x_1)=f(x_2) \implies x_1^3+24x_1+45 = x_2^3+24x_2+45$ then we can simplyfy this by saying that $x_1^3-x_2^3 +24(x_1-x_2) =0$
If $x_1 \neq x_2$ then can not be true,thus they must be equal,thus function is injective thus has inverse
The derivative of $f$ is $$f'(x)=3x^2-18x+33=3(x^2-6x+11)=3[(x-3)^2+2]$$ which is positive. Therefore, $f$ is strictly increasing and hence injective. Nevertheless, an algebraic expression for the inverse it is complicated, because it involves solving of a cubic equation.