Nondifferentiability of $f$ at a point where on the left decreasing and convex, on the right increasing and concave Nondifferentiability of $f$ at a point where on the left decreasing and convex, on the right increasing and concave.
Let $f: \Bbb R\to \Bbb R$ be twice differentiable on $\Bbb R\backslash \{x_0\}$, and
for $x<x_0, f'(x)<0<f''(x)$.
for $x>x_0, f'(x)>0>f''(x)$.
Show $f$ is not differentiable at $x_0$.
Intuitively, there is a cusp at $x_0$, and  $f$ is not differentiable at $x_0$.
But how to prove it analytically?
 A: I will prove that, under the stated assumptions, the following is true:

*

*$\lim_{x\to x_0^-}f'(x)=L$ for some $L\leq 0$.

*Either $\lim_{x\to x_0^+}f'(x)=M$ for some $M>0$ or $\lim_{x\to x_0^+}f'(x)=\infty$
This will show that $f$ is not differentiable at $x_0$ because we will then have that either the left and right-hand derivatives exist but are not equal, or the right-hand derivative is unbounded and thus kills differentiability at $x_0$.
Proof of Result 1: since $f$ is differentiable on $(-\infty,x_0)$, the set $\{f'(x):x<x_0\}$ is non-empty. The assumption that $f'(x)<0$ for every $x\in(-\infty,x_0)$ implies that $0$ is an upper bound for this set, so the supremum property of the reals implies that this set has a least upper bound $L$ satisfying $L\leq 0$. We can now write
$$f'(x)\leq L\text{ for every }x\in(-\infty,x_0)$$
Fix an arbitrary $\varepsilon >0$. Notice that the number $L-\varepsilon$ is not an upper bound of $f'$ over $(-\infty,x_0)$ because $L-\varepsilon$ is less than $L$ and $L$ is, by definition, the least upper bound of $f'$ over this interval. It follows that there is a $\delta>0$ such that $f'(x_0-\delta)>L-\varepsilon$. From the fact that $f''(x)>0$ for every $x\in(-\infty,x_0)$, it follows that $f'$ is strictly increasing on $(-\infty,x_0)$, so for every $x\in\text{dom}[f]$ with $x_0-\delta<x<x_0$, we can write
$$L-\varepsilon<f'(x_0-\delta)<f'(x)\leq L<L+\varepsilon$$
and consequently $|f'(x)-L|<\varepsilon$.
We fixed $\varepsilon>0$ arbitrarily, so the prior argument applies to any positive real number. Thus, for every $\varepsilon>0$, there is a $\delta>0$ such that for every $x\in\text{dom}[f]$ with $x_0-\delta<x<x_0$, $|f'(x)-L|<\varepsilon$. It follows from the definition of a left-hand limit that
$$\lim_{x\to x_0^-}f'(x)=L$$
and we already know that $L\leq 0$. $\blacksquare$
Proof of Result 2: since $f$ is differentiable on $(x_0,\infty)$, the set $\{f'(x):x>x_0\}$ is non-empty. This set either has an upper bound or it doesn't. If it does, then by the supremum property, this set has a least upper bound $M$, so we can write
$$0<f'(x)\leq M\text{ for every }x\in(x_0,\infty)$$
which implies that $M>0$.
For similar reasons as in the proof of Result $1$, fixing an arbitrary $\varepsilon>0$, we can deduce that $M-\varepsilon$ is not an upper bound for $f'$, so for some $\delta>0$,
$$f'(x_0+\delta)>M-\varepsilon$$
Since $f''(x)<0$ for every $x\in(x_0,\infty)$, it follows that $f'$ is strictly decreasing over $(x_0,\infty)$, so for every $x\in\text{dom}[f]$ with $x_0<x<x_0+\delta$,
$$M-\varepsilon<f'(x_0+\delta)<f'(x)\leq M<M+\varepsilon$$
and thus $|f'(x)-M|<\varepsilon$. Since $\varepsilon>0$ was fixed arbitrarily, we can conclude from the definition of a right-hand limit that $\lim_{x\to x_0^+}f'(x)=M>0$.
What if the set doesn't have an upper bound?
Then it follows that $\lim_{x\to x_0^+}f'(x)=\infty$. To see this, fix an arbitrary $K>0$. Since the set $\{f'(x):x>x_0\}$ doesn't have an upper bound, it follows that there is a $\delta >0$ such that $f'(x_0+\delta)>K$. Again from the fact that $f'$ is strictly decreasing, we can deduce that for every $x\in\text{dom}[f]$ with $x_0<x<x_0+\delta$,
$$K<f'(x_0+\delta)<f'(x)$$
so $f'(x)>K$. $K>0$ was fixed arbitrarily, so for every $K>0$, there is a $\delta>0$ such that for every $x\in\text{dom}[f]$ with $x_0<x<x_0+\delta$, $f'(x)>K$. By definition, this is equivalent to $\lim_{x\to x_0^+}f'(x)=\infty$. $\blacksquare$
Proof that $f$ is not differentiable at $x_0$: from Result $2$, either $\lim_{x\to x_0^+}f'(x)=M$ for some $M>0$ or $\lim_{x\to x_0^+}f'(x)=\infty$. If $\lim_{x\to x_0^+}f'(x)=\infty$, then from L'Hopital's Rule,
$$\lim_{x\to x_0^+}\frac{f(x)-f(x_0)}{x-x_0}=\infty$$
so $f$ is not differentiable at $x_0$. If $\lim_{x\to x_0^+}f'(x)=M$ for some $M>0$, then Result $1$ and two applications of L'Hopital's implies that
$$\lim_{x\to x_0^+}\frac{f(x)-f(x_0)}{x-x_0}=M>0\text{ and }\lim_{x\to x_0^-}\frac{f(x)-f(x_0)}{x-x_0}=L\leq 0$$
It follows that the left and right-hand derivatives do not coincide ($L\leq 0 < M$ and consequently $L< M$), so $f$ is again not differentiable at $x_0$. $\blacksquare$
Remark: in the proofs of Results $1$ and $2$, the second derivative was used solely to prove that $f'$ is strictly monotonic on $(-\infty,x_0)$ and $(x_0,\infty)$, so we can weaken the assumptions of the problem to "$f'$ is strictly increasing on $(-\infty,x_0)$" and "$f'$ is strictly decreasing on $(x_0,\infty)$".
