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Qusestion: Let f be a continuous and differentiable function on $[0, \infty[$, with $f(0) = 0$ and such that $f'$ is an increasing function on $[0, \infty[$. Show that the function g, defined on $[0, \infty[$ by $$g(x) = \begin{cases} \frac{f(x)}{x}, x\gt0& \text{is an increasing function.}\\ f'(0), x=0 \end{cases}$$

I have tried to solve this problem but I don't know whether I have done it right or not.

Solution: I have applied mean value theorem on the interval $[0, x]$. Then, $$\frac{f(x)}{x} =f'(c) , 0\lt c \lt x$$

It is given that $f'$ is an increasing function. So I deduce that $\frac{f(x)}{x}$ is also increasing.

Further, $$g(x) = f'(c) \text {such that } 0<c<x$$ Therefore, $$g(0) =f'(c) \text{such that} 0<c<0$$ So, $c=0$

Thus $g(x) = f'(0)$ at $x=0$

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Hm ... you say $f(x)/x = f'(c_x)$ and conclude that $x \mapsto f(x)/x$ is increasing. But why is $x \mapsto c_x$ increasing? –  martini Apr 12 '12 at 8:52

3 Answers 3

up vote 2 down vote accepted

With differentiation we get:$$g'(x)=\frac{f'(x)*x-f(x)}{x^2}$$ We want to show that $g'(x)>0 \ \Leftrightarrow \ f'(x)*x-f(x) > 0 \ (1)\ \forall \ x > 0 $ Reforming (1) we got: $$f'(x) > \frac{f(x)}{x} \ (2)$$ Applying the mean value theorem at [0,x], we got $f'(c)=\frac{f(x)}{x}$, $ 0<c<x$. So we want to show that $f'(x) > f'(c) \ \forall x > 0$. But since f' is an increasing function we got: $$ x > c \Leftrightarrow f'(x)>f'(c)=\frac{f(x)}{x}$$. So from (2) and since $g'(x) > 0 \ \forall \ x >0 $, g(x) is an increasing function in $[0,\infty).$

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$c$ depends on $x$, and what you did doens't prove that if $x_1\leq x_2$ then $c_{x_1}\leq c_{x_2}$.

But we can write for $x>0$, since $f'$ is increasing hence integrable over finite intervals $$\frac{f(x)}x=\frac{f(x)-f(0)}x=\frac 1x\int_0^xf'(t)dt=\int_0^1f'(xs)ds$$ by the substitution $t=xs$. This formula also works for $x=0$, and now it's easy to deduce that $g$ is increasing: if $x_1\leq x_2$, for all $0\leq s\leq 1$ we have $sx_1\leq sx_2$ and since $f'$ is increasing...

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Base on @chemeng $g(x)$is increasing in$(0,\infty)$.when $x=0$, $$\lim_{x \to 0^+} g(x)=\lim\frac{f(x)-f(0)}{x-0}=f'(0)=g(0) $$,so $g(x)$ is contentious at $0$,last g(x) is an increasing function in $[0,\infty)$

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