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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!

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    $\begingroup$ 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
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$lim_{x\rightarrow 3}{{a(x)-a(3)}\over{x-3}}=lim_{x\rightarrow 3}{{(x-3)^2f(x)}\over{x-3}}=lim_{x\rightarrow 3}(x-3)f(x)$. 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 $a'(3)=0$.

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  • $\begingroup$ 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

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