Let $f:[0,1]\to\Bbb{R}$ a continuous function which is differentiable in $(0,1)$, but is not in $0$ and $1$ I need to draw a graphic which the function is differentiable in whole $(0,1)$ but is not in $0$ and $1$.
I just can imagine the function $f(x)=cotg(\frac{x}{\pi})$ but this $f$ is not well-defined in $[0,1]$, despiste being not differentiable at $0$ and $1$.
 A: Oh, I missed the restriction that the domain of $f$ is $[0,1]$.  If $f$ is differentiable on $(0,1)$ and continuous on $[0,1]$ and nonexistent on $(-\infty,0)$ and $(1, \infty)$ this is impossible unless $\lim f'(x) = \pm \infty$.
$f'(0) = \lim_{h\rightarrow 0^+}\frac{f(h) - f(0)}{h}$.
As $f$ is continuous $f(0) = lim_{x\rightarrow 0^+} f(x)$ so
$f'(0) = \lim_{h\rightarrow 0^+}\frac{f(h) - f(0)}{h}= \lim_{x\rightarrow 0^+}\lim_{h\rightarrow x^+}\frac{f(x+h) - f(x)}{h}=\lim_{x\rightarrow 0^+} f'(x)$
So is differentiable at $x = 0$ if the limit exists.  Likewise:
$f'(1) = \lim_{h\rightarrow 0^+}\frac{f(1-h) - f(1)}{-h}$.
As $f$ is continuous $f(1) = lim_{x\rightarrow 1^-} f(x)$ so
$f'(1) = \lim_{h\rightarrow 0^+}\frac{f(h) - f(0)}{h}= \lim_{x\rightarrow 1^-}\lim_{h\rightarrow x^-}\frac{f(x-h) - f(x)}{-h}=\lim_{x\rightarrow 1^-} f'(x)$
So is differentiable at $x=1$ if the limit exist.
So we want a function where $lim f(x)$ exists but $\lim f'(x)$ doesn't.
Graphically a function that goes vertical at $x= 0,1$ will do this. ie.
$f(x) = \sqrt{\frac 14 - (\frac 12 - x)^2}$
[$f'(x) = -\frac 2{\sqrt{\frac 14 - (\frac 12 - x)^2}2 (\frac 12 - x)^2}$ which is defined for $0 < x < 1$ but undefined at $x=0$ or $x=1$]
BUT if $f:\mathbb R \rightarrow \mathbb R$ where $f$ is continuous and $f$ is differentiable on $(0,1)$ but not differentiable at $0$ or $1$ is certainly possible if $f$ "goes off at a sharp angle" at 0 and 1.
Examples: $f(x) = |x(x-1)|$.  for $f(x) = 0; x < 0; f(x) = x; 0 \le x \le 1; f(x) = 1; x > 1$ or $f(x) = x^2; x < 0; f(x)=x; 0 \le x \le 1; f(x)=-x^3 + 2; x > 1$, etc.
In all of these $\lim_{h\rightarrow 0^+}\frac{f(x + h) - f(x)}{h}\ne \lim_{h\rightarrow 0^-}\frac{f(x + h) - f(x)}{h}$ at $x = 0$ or $x=1$ so are not differentiable there.
A: You've already seen
$$
f_{1}(x) = \sqrt{x - x^{2}},
$$
whose graph is the top half of a circle over $[0, 1]$; other examples include suitable scalings of $\arccos$ or $\arcsin$, such as
$$
f_{2}(x) = \arcsin\bigr(\tfrac{1}{2}(1 + x)\bigr),
$$
which again has vertical tangents at the endpoints.
To get more spectacular failure, try something like
$$
f_{3}(x) = f_{1}(x) \sin\left(\frac{1}{f_{1}(x)^{n}}\right),\quad n \geq 1,
$$
extended by continuity to be $0$ at the endpoints. Here, the derivative oscillates (and is unbounded) near each endpoint, so the endpoint derivatives do not exist even if you allow $\pm\infty$ as values.

A: Hint: First, find a continuous function $g$ on $[0,1]$ differentiable everywhere except $0$. Let $h(x)=g(x)\cdot(x-\frac{1}{2})^2$ -- it will have the same properties, but the derivative at $\frac{1}{2}$ will be $0$. Then put $$f(x)=\begin{cases} h(x) \textrm{ if }x<\frac{1}{2}\\ h(1-x)\textrm{ if }x\geq \frac{1}{2}.\end{cases}$$
A: $$f(x) = x \sin \frac {1}{x} + (1-x) \sin \frac {1}{1-x} \;\;\;\;\text{with}\;\; f(0) = f(1) = \sin 1$$
It is straightforward to show that the function is continuous on $[0,1]$, differentiable on $(0,1)$, but is not differentiable at $\{0,1\}$. Its graph can be plotted at Wolfram Alpha.
A: Why not simply $f (x)=x$ if $0 \le x \le 1$.$  f (x)=0 $ if $x <0$ and $f (x)=1$ if $x>1$?
$f $ is continuous and not differentiable at $0$ or $1$.
