I would like to find the inverse function of $\displaystyle f(x)=x^3+3^x$
Since $\displaystyle a^b = \exp{b\ln{a}}$
$\displaystyle y=x^3+\exp{x\ln{3}}$
I do not know how to solve for x in terms of y.
Let $\displaystyle f(x)=x^3+3^x$. show that f has an inverse and find $\displaystyle (f^{-1})'{(4)}$.
$\displaystyle (f^{-1})'{(c)} = \frac{1}{f'(a)}$.
If f(a) = 4, then a = 1. Since $\displaystyle f'(x)=3x^2+3^x\ln{3}$, f'(1)=3+3ln3
Thus,
$\displaystyle (f^{-1})'{(4)} = \frac{1}{f'(1)} = \frac{1}{3(1+\ln{3})} $.
I was just curious to see what the inverse of f was. I assume that to show that f has an inverse means to solve for the inverse. Since this function is an increasing function, it has an inverse.