Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
1  
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

3 Answers 3

up vote 1 down vote accepted

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 $(-x)^3=-x^3$.

share|improve this answer
    
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.

share|improve this answer

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 $(0, \pi]$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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