Showing differentiability of piece-wise function Problem:
Let $f: \mathbb R \to \mathbb R$ be
$f(x)=\begin{cases}\cos(x+\pi), & x>0 \\ x² + x - 1, & x\leq 0 \end{cases}$
Is f differentiable?
Soltuion:
Continous: $\lim_{x\to 0, x>0}f(x)=-1=\lim_{x\to 0, x < 0}f(x)=-1$
The derivative on $\mathbb R\setminus \{0\}$ is
$f(x)=\begin{cases}-\sin(x+\pi), & x>0 \\ 2x + 1, & x\leq 0 \end{cases}$
We have (1) $\lim_{x\to 0, x > 0} f'(x)=0$ and $\lim_{x\to 0, x < 0}f(x) = 1$, so $f$ isn't differentiable at $0$, so f is not differentiable on $\mathbb R$
Question: How does the argument (1) show that f isn't continouse? As far as I am concerned, it only shows that $f'$ isn't continouse at 0 but it doesn't say anything about it's existence.
So how does (1) work for arguing that $f$ isn't differentiable?
 A: If $f$ was differentiable at $0$, the left derivative would be equal to the right derivative (by unicity of the limit). You proved that those numbers aren’t equal. Hence $f$ can’t be differentiable at $0$.
A: $f$ is not differentiable at 0 as you have shown that the left and the right limit for $f'(x)$ is not equal at $x=0$ 
$$\underset{x\to {0}^{-}}{\mathrm{lim}}f'\left(x\right)\neq \underset{x\to {0}^{+}}{\mathrm{lim}}f'\left(x\right)$$
but $f$ is continuous at 0 because the left limit and right limit is equal for $f(x)$ at $x=0$
$$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$$
A: The given function is $f: \mathbb R \to \mathbb R$ 
$f(x)=\begin{cases}\cos(x+\pi), & x>0 \\ x² + x - 1, & x\leq 0 \end{cases}$
A function is differentiable if: (i) it is continuous on that interval (ii)it has no sharp corner or cusp and (iii)the derivative exists(finite) on that interval i.e. has a tangent line with a finite slope.
Now,(i) for $x \gt 0,f(x)=\cos(x+\pi)=-\cos x$ and for $x \le 0,f(x)=x^2+x-1 $.Thus,$f(x \approx 0) \approx -1$.Thus $f$ is continuous.
(ii)for $x \le 0$ the function $f(x-0)=x^2+x-1$  
for $x \gt 0$ the function is $f(x+0)=-\cos x \approx -1+\dfrac{x^2}{2}$ 
the two parabolas  $f(x-0)$ & $f(x+0)$ are not the same(they have different forms),so the function $f$ isn't smooth at $0$
(iii)the derivative $f'$ doesn't exist on any (small) interval containing the origin.
Hence $f$ is not differentiable at $0$.
