I have recently learned the Mean Value Theorem and have been trying to do a few proofs using it. The first question I approached says use the Mean Value Theorem to prove that $\sqrt{1+h}$$<$$ 1 +$ $h\over 2$ for all $h > 0$. My attempt goes like this: Let $h > 0$ and $f(x)$ = $\sqrt{1+x}$. Now, by the Mean Value Theorem, there must exist t $\in (0,h)$ such that $f'(t) =$ $f(h) - f(0) \over h - 0$. This means that $f(h) =$ $f'(t)(h-0) + f(0)$ = $h \over 2\sqrt{1 + t}$ $+ 1 $. Thus, $h \over 2\sqrt{1 + t}$ $<$ $1 +$ $h\over 2$. Is my approach to this problem correct? Are there any improvements that should be made to the proof? Also, another problem asks for $n \geq 1$ and $0 \leq y \leq x$ show that n$y^{n-1}$$(x - y)$$\leq$$x^n$ $-$ $y^n$$\leq$$nx^{n-1}$$(x - y)$ My main problem with this is that it is a function of two variables, so I am not sure how to assign the function which I take the derivative of. Any hints would be much appreciated.


For the second problem :

Take $f(x)=x^n$ so $f'(x)=nx^{n-1}$ .

The inequality is now equivalent with :

$$f'(y) \leq \frac{f(x)-f(y)}{x-y} \leq f'(x)$$

From the mean value theorem there is some $a \in (y,x)$ such that :


The inequality is now equivalent with :

$$f'(y)<f'(a)<f'(y)$$ but because $f'$ is increasing this is equivalent with :

$$y<a<x$$ which is true because $a \in (y,x)$ .

  • $\begingroup$ Thank you very much! After reading this answer the problem is very clear to me $\endgroup$ – Matt Dyer Dec 16 '15 at 18:09

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