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$$h(x)=\begin{cases} {4x+2x^2 \sin \left(\frac{1}{x}\right)}{} & \text{if $x <0$} \\[6pt] {x}\cdot{\cos(x)}+ kx & \text{if $x\geq0$} \\ \end{cases} $$

I know how to prove they are continuous but how do I show the function is differentiable? Any ideas please? I think $k=3$

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Let $f$ be such a function, you may try to expand the details about \begin{align*} \lim_{h\rightarrow 0^{+}}\dfrac{f(h)-f(0)}{h}=\lim_{h\rightarrow 0^{-}}\dfrac{f(h)-f(0)}{h} \end{align*} and solve for $k$.

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  • $\begingroup$ Can't I differentiate both sides and put x=0 to find k? $\endgroup$ – nerv21 Dec 5 '17 at 19:48
  • $\begingroup$ I guess you mean $\lim_{x\rightarrow 0^{+}}f'(x)=\lim_{x\rightarrow 0^{-}}f'(x)$, in this case, it works, but it needs some reasoning, but the safest way to do this question is the left-right limit of the difference quotient. $\endgroup$ – user284331 Dec 5 '17 at 19:51
  • $\begingroup$ You can differentiate the expression on the right and plug in $0$, but that won't work for the expression on the left, which is not defined at $0$. And taking the limit of $f'(x)$ as $x \to 0^+$ has no guarantee of being correct. (Contrary to what @user284331 says, in this case it does NOT work. (I expect that user284331 just didn't check carefully; this shouldn't be controversial.)) $\endgroup$ – Toby Bartels Dec 5 '17 at 19:57
  • $\begingroup$ (If you use, say, $4x$ instead of $4x + 2x^2 \sin(1/x)$ on the left, then it's OK to differentiate and plug in $0$, because even though that formula is only given to apply when $x < 0$, the fact that it has a derivative if extended to (and even beyond) $0$ means that the derivative that you get this way is the desired left-hand limit. But $4x + 2x^2 \sin(1/x)$ doesn't give you any result at $x = 0$.) $\endgroup$ – Toby Bartels Dec 5 '17 at 20:00
  • $\begingroup$ @TobyBartels, I am sorry that I didn't check carefully, yes, the left side would cause some problem. So that's why I said the safest way is to deal with the both-sided difference quotient. $\endgroup$ – user284331 Dec 5 '17 at 20:03
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Let $\Delta (x)=\frac{f (x)-f (0)}{x-0} $

with $f (0)=0.$

$$\lim_{0^-}\Delta (x)=\lim_{0^-}(4+2x\sin (1/x))=4$$ since $$|x\sin (1/x)|\le |x|$$

on the right,

$$\lim_{0^+}\Delta (x)=\lim_{0^+}(\cos (x)+k)=1+k $$

$f $ is differentiable at $0$ if $$1+k=4. $$

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