# Using Mean Value Theorem to Generalize

Using Mean Value Theorem, prove that:

$$\sqrt{1+h}<1+\frac{1}{2}h \text{ for } h>0.$$

Now, generalize this as follows: If $$0 and $$h>0,$$ then show that

$$(1+h)^p<1+ph.$$

Assume the usual rules about differentiating powers.

I have no idea, how to even start this. I can state the mean value theorem, but it's hard to see how that's used as there is no explicit function stated.

MVT: If $$f:[a,b]→\mathbb R$$ is continuous on $$[a,b]$$ and differentiable on $$(a,b)$$, Then there is a $$c\in (a,b)$$ such that $$f'(c)=\frac{f(b)-f(a)}{b-a}$$

I suppose I could generate functions, say, $$f(x)=\sqrt{1+x}, g(x)=1+\frac{1}{2}x, m(x)=\sqrt{1+x}-(1+\frac{1}{2}x)=f(x)-g(x)$$. I want to show that $$m(x)<0$$ for $$x>0$$?

• Consider $f(x)=x^p$ and estimate the derivative. Nov 12, 2019 at 20:02

$$\frac{(1+h)^p-1}{h} = \frac{p}{(1+\xi)^{1-p}} < p$$
where $$0<\xi.
You can use $$f$$, where taking $$x=h$$ and $$x=0$$, for the mean value th. you have that exists $$c\in (0,h)$$ such that $$\sqrt{1+h} - 1 = f(h)-f(0) = hf'(c)$$. The next step is to dimension $$f'(c)$$, you do it!