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Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a differentiable function such that $f'$ is increasing. Setting $a\in\mathbb{R}$, prove: the function $g(x)=\frac{f(x)-f(a)}{x-a}$ is also an increasing function on $(a,\infty)$ and $(-\infty,a)$

My thoughts:

By MVT: $\exists c\in(a,x_0) \ s.t. f'(c)=\frac{f(x_0)-f(a)}{x-a}, \ \exists d\in(x_0,x_1) \ s.t. \ f'(d)=\frac{f(x_1)-f(x_0)}{x_1-x_0}$

$d>c \Rightarrow f'(d)\geq f'(c)$

Now I got stuck since I couldn't show $\frac{f(x_1)-f(a)}{x_1-a} \geq \frac{f(x_1)-f(x_0)}{x_1-x_0}$.

Any help appreciated.

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Note that $g$ is differentiable when $x\ne a$. Thus, we only need to show that $$g'(x)=\frac{f'(x)(x-a)-(f(x)-f(a))\cdot1}{(x-a)^2}\ge 0,$$ or $$f'(x)(x-a)\ge f(x)-f(a).$$ If $x>a$, we only need to show that $$f'(x)\ge\frac{f(x)-f(a)}{x-a}.$$ By MVT, the RHS is $f'(c)$ for some $c\in[a,x]$. And clearly $$f'(x)\ge f'(c).$$

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  • $\begingroup$ Could you please explain why $g$ is differentiable when $x\neq 0$ and how the inequality follows from it? $\endgroup$ – Itay4 Mar 7 '17 at 11:01
  • $\begingroup$ First, $g$ is the quotient of two differentiable function; hence differentiable. Second, use MVT. $\endgroup$ – Eclipse Sun Mar 7 '17 at 11:04
  • $\begingroup$ But why is it not differentiable when $x=0$? And I can't understand how that shows $g$ is increasing. $\endgroup$ – Itay4 Mar 7 '17 at 11:10
  • $\begingroup$ That's a typo. I mean $x\ne a$. $\endgroup$ – Eclipse Sun Mar 7 '17 at 11:13
  • $\begingroup$ Ok, that make sense. But I still can't see why that proves the $g$ is increasing.. $\endgroup$ – Itay4 Mar 7 '17 at 11:18

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