Given is a (not necessary continuous) function $f: \mathbb{R} \rightarrow \mathbb{R}$ with $|f(x)|<1$ for all $x \in \mathbb{R}$. Let $a(x):=(x-3)^{2}f(x)$. Prove that $a$ is differentiable at $x_{0}=3$.

We know from a previous task that $b(x)=(x-3)f(x)$ is continuous at $x_{0}=3$

From this we can conclude that this function $a$ will be continuous as well because stuff like summation or multiplication of continuous things stay continuous.

Now we know that $a(x)$ is continuous but what does it tell us? Unfortunately, a continuous function isn't necessarily differentiable...

What to do here? I thought about using the difference quotient but we will have troubles with $f(x)$, I mean it could be smaller than $0$ if we don't use the modulus. Or can we just set the modulus when we use difference quotient?

This is no homework, it's from an old exam and if you want I can upload it here but it won't be in English!

  • 1
    $\begingroup$ Maybe you should try the easier problem where you replace 3 by 0. The difference quotient is a good idea. $\endgroup$ – Matthew Towers Oct 8 '16 at 11:27
  • $\begingroup$ Am I allowed to just set the modulus for $f(x)$ when I use the difference quotient? That's what I wasn't sure about. $\endgroup$ – cnmesr Oct 8 '16 at 11:37

$lim_{x\rightarrow 3}{{a(x)-a(3)}\over{x-3}}=lim_{x\rightarrow 3}{{(x-3)^2f(x)}\over{x-3}}=lim_{x\rightarrow 3}(x-3)f(x)$. This implies that $|lim_{x\rightarrow 3}{{a(x)-a(3)}\over{x-3}}|\leq lim_{x\rightarrow 3}|x-3|=0$. So $a$ is differentiable at $3$ and $a'(3)=0$.

  • $\begingroup$ Haha you again, thanks so much! :-)) $\endgroup$ – cnmesr Oct 8 '16 at 11:43
  • $\begingroup$ I feared it's going to be as complicated as here: math.stackexchange.com/questions/269666/… But actually it's that simply done, awesome. $\endgroup$ – cnmesr Oct 8 '16 at 11:46

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.