Differentiability and monotoncitiy

Question

Prove or give an opposite example:

Let $x_0 \in \mathbb{R}$,

$f:\mathbb R \to \mathbb R$ differentiable with $f'(x_0)=0$ and $ f'(x)\gt 0$ for all $x\neq x_0$. Then $f$ is strictly monotonic on $\mathbb{R}$.

I'm having trouble understanding how to solve this. I know for a monotonic function:

$\forall a,b \in \mathbb R$, $a\lt b \implies f(b)\gt f(a)$.

I'd like a hint to help solve this question

Answer 1

Because $f'(x) > \forall x \gt x_0$ then $f$ is strictly increasing 0n $(x_0, \infty)$. Now, because $f$ continuous it follows $f$ is strictly increasing on $[x_0, \infty)$. Similar, $f$ is strictly increasing on $(-\infty, x_0]$ therefore $f$ is strictly increasing on $(-\infty, \infty)$

Answer 2

HINT: Mean Value Theorem. (Consider $x\le x_0$ and $x\ge x_0$ separately.)

Answer 3

Prove the following with the mean value theorem, given that $x<y$:

• $f(x)\leq f(y)$
• if $f(x)=f(y)$ then $x_0\in(x,y)$
• $f(x)<f(\frac{x+y}{2})\leq f(y)$ or $f(x)\leq f(\frac{x+y}{2})<f(y)$

Answer 4

A more general claim is true:

Claim. Let $f\colon \Bbb R\to\Bbb R$ be differentiable such that $f'(x)\ge 0$ for all $x\in \Bbb R$. Also assume that between any two points, there is at least one point with non-zero derivative. Then $f$ is strictly increasing. (So in principle many more than just one point is allowed to have zero derivative)

Proof. Let $a,b\in \Bbb R$ with $a<b$. We want to show that $f(a)<f(b)$. By assumption, there exists $c$ with $a<c<b$ and $f'(c)>0$. By definition of derivative, we have $\frac{f(c+h)-f(c)}h\approx f'(c)$ for small $h$, or more precisely and in particular: For all sufficiently small $h>0$, we have $\left|\frac{f(c+h)-f(c)}h-f'(c)\right|<\frac12|f'(c)|$ and therefore $f(c+h)>f(c)$. We may assume that $h<b-c$. So with $d=c+h<b$ we have $f(d)>f(c)$. By the intermediate value theorem, $f(c)-f(a)=(c-a)f'(\xi)$ and $f(b)-f(d)=(b-d)f'(\eta)$ with $a<\xi<c$ and $d<\eta<b$. This implies $$f(b)\ge f(d)>f(c)\ge f(a)$$ as desired.