Prove (x^5)+(e^x)=0 has at most 1 real root

Question

I know to suppose (x^5)+(e^x)=0 has two distinct roots. f(a)=f(b)=0 with a/=b , there exists a c E(a,b). Now I am stuck. 

Answer 1

if f(x) = x⁵+e^x

then f'(x) = 5x⁴+e^x

 

since f'(x) is positive for all real numbers (we know this because x4 and e^x are both always positive), that means that f(x) always has a positive slope.  In order for f(x) to have more than one root, it MUST at some point "turn around" to hit the x axis twice.  That means that it has to have a positive slope, then a negative slope (or vice versa).  Since we know that can't happen, we know that f(x) can only have AT MOST 1 real root