0
$\begingroup$

Having problem with it, but first I'll explain the question.

I'm asked to prove that $g(x)=|x|f(x)$ is derivable in $x_0=0$ iff $f(0)=0$, while $f$ is a continuous function $x_0=0$.

Basically since $f(x)=|x|$ is not derivable in $x=0$, it might seem like a problem, but I think that the definition of whether a function is derivable or not is that if its limit exists, i.e: $f'(x_0) = \lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}$. so basically since $|x|$ changes the pluses to minuses at times, the only opportunity where the definition will hold is when $x=0$. Is it true? and how should I write it mathematically?

thank you in advance

$\endgroup$
  • 1
    $\begingroup$ (if:) $\lim_{x\to0}\frac{g(x)-0}{x}=\lim_{x\to0}\text{sgn}(x)f(x)=0$, where the last limit is because $f\to0$ and $\text{sng}$ is bounded. $\endgroup$ – OR. Jun 27 '17 at 12:29
3
$\begingroup$

$g $ differentiable at $x=0 \iff$

$$ \lim_{x\to 0^+}\frac {g (x)-g (0)}{x}=\lim_{x\to 0^-}\frac {g (x)-g (0)}{x}$$ $$\iff \lim_{x\to0^+}\frac {xf (x)-0}{x}=\lim_{x\to0^-}\frac {-xf (x)}{x} $$ $$\iff f (0)=-f (0)=0$$

$\endgroup$
0
$\begingroup$

compute the diffrerential Quotient: $$\frac{|x_0+h|f(x_0+h)-|x_0|f(x_0)}{h}$$ for $$x_0=0$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.