$f(0)=0, \ f'(X)$ is a strictly increasing function: prove that $f(X)/X$ is also strictly increasing for all $X \in(0, \infty)$
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Could you please define MSIF, and show what work you've done on the problem? – Adrian Keister Jul 30 '18 at 18:50
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MSIF is monotonically strictly increasing function . – prakhar Gupta Jul 30 '18 at 18:55
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I assume f(X)/X as H(X) and differentiate it H'(X) but could able to proof that H'(X) is greater than 0 – prakhar Gupta Jul 30 '18 at 18:57
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What arithmetic error? – prakhar Gupta Jul 30 '18 at 19:33
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3Duplicate of Show that $x \mapsto \frac{f(x)}{x}$ is strictly increasing on (0,1) given that f '(x) is strictly increasing on (0,1) and that f(0)=0. Also of Show that if $f'$ is strictly increasing, then $\frac{f(x)}{x}$ is increasing over $(0,\infty)$. – dxiv Jul 30 '18 at 19:37
3 Answers
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For any $X>0$, $f$ is continuous in $[0,X]$ and differentiable in $(0,X)$. By mean value theorem, there exists $c\in (0,X)$ such that $f'(c)= \frac {f(X)-f(0)} {X-0}=\frac {f(X)} {X}$ Since $f'$ is strictly increasing and $X>c$, we have $f'(X) > f'(c)$, i.e. $ X f'(X) - f(X)>0$
Consider the function $g(X)=\frac {f(X)} X$ $$g'(X)= \frac 1 {X^2} (X f'(X) - f(X))>0$$ So derivative of $g$ is strictly positive. So, $g$ is strictly increasing function.
Vignesh
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Hint: Apply the mean value theorem to $f$ on the interval $[0, x]$, for $x > 0$.
youngsmasher
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Alright, so we've got 2 conditions:
- $f(0) = 0$
- $f'(x)$ is strictly increasing
Define $h(x) = \frac{f(x)}{x}$. Then as you suggested, taking the derivative yields: $$h'(x) = \frac{x f'(x) - f(x)}{x^2}$$
We want to show that $h(x) > 0$ for all $x \in (0, \infty)$. To see this, note that by the fundamental theorem of calculus and condition $1$ we have: $$f(x) = \int_{0}^x f'(t)dt$$ Then by the mean value theorem, we have that there is some $c \in (0,x)$ such that: $$ \int_{0}^x f'(t)dt = xf'(c)$$ By condition $2$, since $c < x$ we have $f'(c) < f'(x)$. Combining this with the above yeilds: $$f(x) < xf'(x)$$
And now we can complete our proof! The above inequality gives us that $xf'(x) - f(x) > 0$, and for any positive $x$ we know that $x^2 > 0$. Thus, both the numerator and denominator of $h'(x)$ are positive, and therefore $h'(x) > 0$ for all $x \in (0, \infty)$.
Therefore $\frac{f(x)}{x}$ is strictly increasing.
Joe
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