if $|f''(x)|\le |f'(x)|,\forall x\in R$ show that either $f'(x)>0$ or $f'(x)<0,\forall x\in R$ Edit:let $f:R\to R$ be twice differenriable on $R$, such that
$$|f''(x)|\le |f'(x)|,\forall  x\in R$$
show that either:
$f'(x)>0$
or $f'(x)<0,\forall x\in R$
since
$$0\ge (f''(x))^2-(f'(x))^2=(f''(x)-f'(x))(f''(x)+f'(x))$$
so
$$f"(x)+f'(x)\ge 0,\mbox{and} f''(x)-f'(x)\le 0$$
or
$$f"(x)+f'(x)\le 0,\mbox{and} f''(x)-f'(x)\ge 0$$
 A: let $f(x)=-2x$. One has  $f'(x) = -2$ and $f''(x) = 0$ for all $x$. We see that $|f''(x)|\le |f'(x)|$ for all $x$ but $f'(x) = -2<0$ for all $x$.
A: Let $g=f'$. Then $g$ is once differentiable, we have $|g'| \leq |g|$
on $\mathbb R$, and we wish to show that $g$ always keeps the same sign.
Suppose that $g$ is nonzero, so that there is a $x_0\in{\mathbb R}$ with
$g(x_0)\neq 0$. Replacing $g$ with $x\mapsto g(x_0+x)$, we may assume without loss that $x_0=0$, so that $g(0)\neq 0$.
Replacing $g$ with $-g$, we may further assume that $g(0) \gt 0$.
Let $$A=\bigg\lbrace x \gt 0 \ \bigg| \ \forall t\in [0,x], \ g(t) \gt 0 \bigg\rbrace \tag{1}$$.
Then $A$ is a nonempty subset of $[0,\infty)$. Let $a=\sup(A)$ (note that $a$ can be $\infty$). By the construction of $A$, we have
$$[0,a) \subseteq A \tag{2}$$
For $t\in [0,a)$, we then have $-1 \leq \frac{g'(t)}{g(t)} \leq 1$, and integrating, we deduce $-t \leq \ln(g(t))-\ln(g(0)) \leq t$, whence
$g(t) \geq g(0)e^{-t} \ (t\in[0,a]) \label{3}\tag{3}$
If $a$ were finite, by \eqref{3} above and the continuity of $g$ we would have
$g(a) \geq g(0)e^{-a}$, so that $g$ would be positive on a neighborhood of $a$,
contradicting the definition of $a$. So $a=\infty$, and hence $g(t) \geq g(0)e^{-t}$ for all $t\geq 0$. A symmetric argument shows that $g(t) \geq g(0)e^{t}$ for all $t\leq 0$. This finishes the proof.
A: It is just an application of the Gronwall's lemma. Since $\left|f^{\prime}\left(x\right)\right|>\left|f^{\prime\prime}\left(x\right)\right|$ then $\left|f^{\prime}\left(x\right)\right|>f^{\prime\prime}\left(x\right)$. Now, assume there are $a,b\in\mathbb{R}$ such that $f^{\prime}\left(a\right)=0$ and $f^{\prime}$ is positive in $\left(a,b\right)$ Hence, for every $x\in\left[a,b\right)$ we have $f^{\prime}\left(x\right)>f^{\prime\prime}\left(x\right)$ and so, by the Gronwall's lemma, $$f^{\prime}\left(x\right)\leq f^{\prime}\left(a\right)e^{x}=0,\,\forall x\in\left[a,b\right)$$ which is a contradiction. The other cases are similar.
