If f(x) = begin{cases} e^x + a & text{for } x | vedclass

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(C) For $f(x)$ to be differentiable at $x = 0$,it must first be continuous at $x = 0$. For continuity at $x = 0$: $\lim_{x 	o 0^-} f(x) = \lim_{x 	o

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$-4$

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(C) For $f(x)$ to be differentiable at $x = 0$,it must first be continuous at $x = 0$. For continuity at $x = 0$: $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$ $\lim_{x 	o 0^-} (e^x + a) = e^0 + a = 1 + a$ $\lim_{x 	o 0^+} (x - 3) = 0 - 3 = -3$ Equating these,$1 + a = -3 \Rightarrow a = -4$. Now,check for differentiability at $x = 0$ using the left-hand derivative $(LHD)$ and right-hand derivative $(RHD)$: $LHD = \lim_{h 	o 0} \frac{f(0 - h) - f(0)}{-h} = \lim_{h 	o 0} \frac{(e^{-h} - 4) - (-3)}{-h} = \lim_{h 	o 0} \frac{e^{-h} - 1}{-h} = 1$ $RHD = \lim_{h 	o 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h 	o 0} \frac{(h - 3) - (-3)}{h} = \lim_{h 	o 0} \frac{h}{h} = 1$ Since $LHD = RHD = 1$,the function is differentiable at $x = 0$ when $a = -4$.

If $f(x) = \begin{cases} e^x + a & \text{for } x < 0 \\ x - 3 & \text{for } x \geqslant 0 \end{cases}$ is differentiable at $x = 0$,then $a$ equals:

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