Question on differentiability Let f : [a, b] → R be diﬀerentiable, and f(a)=0. Suppose there is a real
number c such that |f'(x)| ≤ c|f(x)| for all x ∈ [a, b]. Prove that f(x) = 0 for
all x ∈ [a, b].
I really Need help for this exercise please. Can someone help me?
 A: This follows from Gronwall's Inequality (see http://en.wikipedia.org/wiki/Gronwall%27s_inequality)
EDIT: it just occurred to me that because you have $|f'(x)|$ rather than $f'(x)$, you may need to do a little work before you apply Gronwall's inequality: if $f$ is not identically zero, there exists $x_0\geq 0$ and $\epsilon > 0$ such that $f(x_0)=0$ but $f(x) \neq 0$ for all $x \in (x_0,x_0+\epsilon)$.  If $f > 0$ on $(x_0,x_0+\epsilon)$, apply Gronwall's inequality and get a contradiction.  If $f < 0$ on $(x_0,x_0+\epsilon)$, apply Gronwall's inequality to $-f$ and get a contradiction.
The link proves something more general.  Note that only basic calculus is needed for the proof.  If you are not allowed to cite this inequality, just mimic the proof in the link.
EDIT: It occurs to me that your question may be a duplicate.  If you search for questions using "Gronwall's Inequality", you may find something useful.
A: Let $n$ big enough s.t. $c(b-a)<n$. At first, we just consider $f(x)$ in $\left[a,a+\dfrac{b-a}{n}\right]$. Assume that $|f(x)|$ reach its maximum at $x=\xi$ in $\left[a,a+\dfrac{b-a}{n}\right]$. Then $$\left|\frac{f(\xi)-f(a)}{\xi-a}\right|=|f'(\nu)|\leqslant c|f(\nu)|,$$where $\nu\in (a,\xi)$. Therefore $$|f(\xi)|\leqslant c(\xi-a)|f(\nu)|\leqslant \frac{c(b-a)}{n}|f(\nu)|\leqslant \frac{c(b-a)}{n}|f(\xi)|.$$Notice that $\dfrac{c(b-a)}{n}<1$, hence we have $|f(\xi)|=0$, thus $f(x)\equiv 0$ in $\left[a,a+\dfrac{b-a}{n}\right]$(from the definition of $|f(\xi)|$ ). Now we can do the same thing in$$\left[a+\dfrac{b-a}{n},a+\dfrac{2(b-a)}{n}\right],\cdots,\left[a+\dfrac{(n-1)(b-a)}{n},b\right]$$
A: Since $f$ is a differentiable function. Consider a fucntion 
$g(x)=(x-b)^n f(x)$
now $g(a)=0=g(b)$ and g satisfies the conditions of Rolle's theorem. This implies that there exists $d$ such that 
$a<d<b$ and $g'(d)=0$
after putting all this and simplifying ,
$f'(d)=(\frac{n}{b-d})f(d)$
Using the inequality given we can see that 
$\frac{n}{b-d}|f(d)| \leq c |f(d)|$
Assuming $|f(d)| \neq 0$
$\frac{n}{b-d} \leq c$
if $c$ is negative then we reach a contradiction by taking $n>0$ this implies $f(d)=0$
if $c$ is positive we can take a value of $n>c$ and reach a contradiction this implies $f(d)=0$
Thus $f$ is identically zero
