# Find The value of $a,b,c$.

If $f(x)=a|\sin x|+be^{|x|}+c|x|^3$ is differentiable at $x=0$, find the values of $a,b,c$.

I know that the derivative exists at $x=0$ iff $f'(0^+)=f'(0^-)$, but I can't find $f'(x)$. Please help. Thanks in advance.

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The question suggests there is only one possible set of values for $a$, $b$, and $c$, so you can reverse-engineer the answer on that basis. –  Dan Brumleve Dec 12 '12 at 7:33
There is not a unique set of $a,b,c$, e.g. $0,0,c$ will do for any $c\in \mathbb R$. –  P.. Dec 12 '12 at 7:43

Hint: Deal separately with positive and negative $x$. To deal with positive $x$, replace $|x|$ everywhere by $x$, and differentiate as usual.

If $x$ is negative, then $|x|=-x$. To deal with negative $x$, replace $|x|$ everywhere by $-x$, and differentiate as usual. It will make life simpler if you note that $\sin(-x)=-\sin x$ and $(-x)^3=-x^3$.

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Thank you.Now I get it. –  Argha Dec 12 '12 at 14:47

$f'(0^-)=\displaystyle{\lim_{x\to0^-}\dfrac{f(x)-f(0)}{x}=\lim_{x\to0^-}\dfrac{a|\sin(x)|+be^{|x|}+c|x|^3-b}{x}\displaystyle}=\\\displaystyle{\lim_{x\to0^-}\dfrac{-a\sin(x)+be^{-x}-cx^3-b}{x}=g'(0)}$
where $g(x)=-a\sin(x)+be^{-x}-cx^3.$
Now in a similar way find $f'(0^+)$ and set $f'(0^+)=f'(0^-)$ to find the relation between $a,b,c$ s.t. $f$ is differentiable.

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Hint: We have, for $x \in [-\pi, \pi]$ (which is enough for this problem):

$$f(x) = \begin{cases}a \sin(x) + be^x + cx^3,\ \mathrm{if}\ x \ge 0 \\ -a \sin(x) + be^{-x} - cx^3,\ \mathrm{if}\ x < 0\end{cases}, x \in [-\pi, \pi]$$

Then differentiate on $[-\pi, 0)$ and $(0, \pi]$.

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What you wrote is not the piecewise version (sinx).. –  Josh Dec 12 '12 at 7:56
@Josh: Fixed. It does work for $x \in [-\pi, \pi]$ :) –  mike4ty4 Dec 13 '12 at 4:22