How to show a function f has at most one real root?

Question

f(x)=x^7+3x+16

What I have is :

Let f(a)=f(b)=0 for all a,b \in R. Then by MVT, there exists c \in (a,b)such that f'(c)=0

f'(c)=f(b)-f(a)/b-a=0

and Im stuck. My professor told me that im on the right track we have to show f(x) cannot be equal to zero at the very end.

Answer 1

What is missing from your proof is the fact that the derivative f'(x)=7x^{6}+3 which can never be 0. Thus, f can not have more than 1 zeros since this would imply that f' also has at least one zero.