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If

$$f(x)=\begin{cases}e^x+a \sin x& \text{ if } x<0\\ b(x-1)^2+x-2 & \text{ if } x\ge 0\end{cases}$$

Then find the values of $a$ and $b$ given that $f(x)$ is differentiable at $x=0$

I worked out to find that $b=3$. Then using the definition of derivibility of function i get

$Rf'(0)=-5$ and $$Lf'(0)=\lim_{h\rightarrow 0} {{e^{-h}+a\sin(-h)}\over {-h}}$$

so how do i solve $LF'(0)$ ??

vadim123
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Aman Mittal
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    I think your numerator should be $e^{-h}-1+a\sin(-h)$. Also, you need to take the limit as $h\rightarrow0^+$ here. – David Mitra Jul 06 '13 at 16:18
  • If you differentiate the $x < 0$ part, you get $e^x +a\cos x$. Plug in $x = 0$ and solve for $a$. – Daniel Fischer Jul 06 '13 at 16:22
  • @DavidMitra HECK !!, I missed out that $-1$..silly me. so $Lf'(0) =a+1$ Thanks David – Aman Mittal Jul 06 '13 at 16:23
  • You're welcome. I presume the finding the limit poses no difficulty now? – David Mitra Jul 06 '13 at 16:27
  • Not at all, Could you look at this one for me please http://math.stackexchange.com/questions/437563/checking-differentiability-for-given-function :) – Aman Mittal Jul 06 '13 at 16:28
  • @DavidMitra We have a continuous function that's left-differentiable in $0$ and right-differentiable in $0$. If the left and right derivatives coincide, the function is differentiable in $0$, that follows directly from the definitions. – Daniel Fischer Jul 06 '13 at 16:31
  • @DanielFischer Ah, sorry. I wasn't thinking clearly before... – David Mitra Jul 06 '13 at 16:35

2 Answers2

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To begin with, it must be

$$\lim_{x\to 0^-}f(x)=\lim_{x\to 0^+}f(x) :$$

$$\lim_{x\to 0^-}f(x)=\lim_{x\to 0^-}(e^x+a\sin x)=1\;\;,\;\;\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}\left(b(x-1)^2+x-2\right)=b-2$$

Thus, it must be that $\;b-2=1\implies b=3\;$ . Now, it also must be

$$f'(0)_-:=\lim_{x\to 0^-}\frac{f(x)-f(0)}x=\lim_{x\to 0^+}f(x)=:f'(0)_+ : $$

$$f'(0)_-:=\lim_{x\to 0^-}\frac{f(x)-f(0)}x=\lim_{x\to 0^-}\frac{e^x+a\sin x-1}x=e^0+a\cos 0=1+a$$

$$f'(0)_+:=\lim_{x\to 0^+}\frac{f(x)-f(0)}x=\lim_{x\to 0^+}\frac{3(x-1)^2+x-2-1}x=-5$$

and thus you get the value of $\,a\,$ and etc.

DonAntonio
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The left hand derivative should be written like this: $$Lf'(0) = \lim_{h\to0^+} \frac{ f(-h) - f(0) }{-h} = \lim_{h\to0^+} \frac{e^{-h} + a \sin(-h) - 1}{-h}.$$

We can then split this into the sum of two convergent limits:

$$Lf'(0) = \lim_{h\to0^+} \frac{e^{-h}-1}{-h} + a\lim_{h\to 0^+} \frac{\sin(-h)}{-h}$$ $$= \lim_{h\to0^+} \frac{e^{-h}-1}{-h} + a\lim_{h\to 0^+} \frac{\sin(h)}{h}$$

Now recall that $\lim_{h\to0^+} \sin(h)/h = 1$. Also notice that the first limit, $ \lim_{h\to0^+} \frac{e^{-h}-1}{-h}$, is the definition of the left hand derivative of $e^x$ at 0. Hence: $$Lf'(0) = 1 + a.$$

Joel
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