Mean Value Theorem x(0 , +infinity) (1+x)^a>1+ax.

Question

I must show that $$(1+x)^a>1+ax ..ifa>1$$ $$(1+x)^a<1+ax ..if..0<a<1$$ $$x(0,+∞)$$ using mean value theorem.What i have done so far. $$(1+x)^a-1-ax>0 ..if..a>1$$ $$(1+x)^a-1-ax<0 ..if..0<a<1$$ using mvt $$f'(x)={f(b)-f(a) \over b-a}$$ $$f(x)=(1+x)^a-1-ax$$ $$f'(x)=a(1+x)^a-1-1$$ $$f(a)=f(0)=1^a-1$$ I cannot understand b value is gone be f(b)=?

Answer 1

$f(x) = x^a.$

1) Let $a>1:$

$\dfrac{f(1+x) -f(1)}{x}= f'(c)$, $1 <c <1+x$.

$(1+x)^a = 1 + x (ac^{a-1} ) > 1+ ax$ ,

since $c^{a-1} > 1^{a-1} =1;$

Can you take it from here?