True or False? #5

A. (counterexample)

$f(x)=|x|,$ then $f(x^2)=x^2.$

$x^2$ is differentiable over all real numbers, while $|x|$ is not.

$\therefore~\boxed{\text{FALSE}}.$

---

B.

$f(x)=0$ is a possible option.

For all $f(x)\neq0,$ $\dfrac{f'(x)}{f(x)}=1,$ and integrating both sides yields $\ln|f(x)|=x+C.$

Since $f(x)=\pm e^{x+C}=\pm e^C \cdot e^x,$ letting $\pm e^C=a,$ we get $f(x)=a\cdot e^x.$ $(a\neq 0,~f(x)\neq0)$

And from that $e^x$ approaches to $0$ if and only if $x$ approaches to negative or positive infinity, we can infer that $f(x)$ cannot be zero for any $x.$

Therefore $f(x)$ can be expressed as a form of $f(x)=a\cdot e^x.$ (Including $f(x)=0,$ since $a=0$ yields $f(x)=0.$ )

$\therefore~\boxed{\text{TRUE}}.$

---

C. (counterexample)

$f(x)=x^{\frac{1}{3}}$ yields $f'(x)=\dfrac{1}{3x^{\frac{2}{3}}},$ and since $f'(0)$ cannot be defined, $f(x)$ is not differentiable at $x=0.$

$\therefore~\boxed{\text{FALSE}}.$

---

From above, we figured out that only B is true.