If $f \neq 0$ there exist $x_0$ with $|f(x_0)| < |f'(x_0)|$ I am facing the following problem. 

Let $f$ be a continuous and derivable function from $[0, 1]$ to $R$,
  such that $f(0) = 0$. Show that if $f \neq 0$ then there exist $x_0$
  such that $|f'(x_0)| > |f(x_0)|$,

I have been thinking for a while on this problem but I have no idea how to demonstrate it. Can someone provide a hint?
 A: Proof assuming that Fundamental Theorem of Calculus can be applied: we have $f(x)=f(x)-f(0)=\int_0^{x} f'(t)\, dt$. If possible let $|f'(x)| \leq |f(x)|$ for all $x$. Then we get $|f(x)|=|\int_0^{x} f'(t)\, dt| \leq \int_0^{x} |f(t)|\, dt$. Integrate both sides and switch the integrals on RHS to get $\int_0^{1} |f(x)|\, dx \leq \int_0^{1}\int_0^{x} |f(t)|\, dt \, dx =\int_0^{1}\int_t^{1} |f(t)|\, dx \, dt=\int_0^{1}(1-t)|f(t)|\, dt$. Using the fact that $1-t <1$ on $(0,1)$ you can easily infer that $f$ must vanish identically. 
Proof without Fundamental Theorem of Calculus:
Suppose $|f'(t)| \leq |f(t)|$ for all $t$. Then $|f(x)|=|f(x)-f(0)| =|x||f'(t)|$  for some $t$ betqween $0$ and $x$. Hence $|f(x)| \leq x |f(t)|$.  Let $c<1$ and let $M$ be the maximum of $|f|$ on $[0,c]$. This maximum is attained at some point $s \in [0,c]$. We  get, by MVT, $M=|f(s)|=|f(s)-f(0)| \leq s|f'(t)|\leq  s|f(t)|\leq cM$ for some $t$ between $0$ and $s$ . Since $c<1$ this implies $M=0$. Hence $f \equiv 0$ in $[0,c]$ for every $c <1$. In turn this implies that $f \equiv 0$ in $[0,1]$.
A: Since $|f|$ is continuous on the closed interval $[0,1]$, $|f|$ assumes a maximum value, $b$ say.

Since $f\ne 0$, it follows that $b > 0$.

Let $S=\{x\in [0,1]{\;:\;}|f(x)|=b\}$.

Since $|f|$ is continuous, $S$ is closed.

Since $S\subseteq [0,1]$, $S$ is compact, hence has a least element, $a$ say.

Since $0\not\in S$, we have $a>0$.

Since $f$ is differentiable, applying the Mean Value Theorem, there exists $c\in (0,a)$ such that
$$f'(c)=\frac{f(a)-f(0)}{a-0}=\frac{f(a)}{a}$$
hence
$$|f'(c)|=\left|\frac{f(a)}{a}\right|=\frac{|f(a)|}{|a|}=\frac{b}{|a|}\ge \frac{b}{1}=b > |f(c)|$$
