The function f(x) = begin{cases} e^x + ax, & | vedclass

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

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$a = -3, b = 1$

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(C) For $f(x)$ to be differentiable at $x = 0$,it must first be continuous at $x = 0$.Left-hand limit $(LHL)$: $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} (e^x + ax) = e^0 + a(0) = 1$.Right-hand limit $(RHL)$: $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} b(x - 1)^2 = b(0 - 1)^2 = b$.Since $f(x)$ is continuous,$LHL = RHL$,so $b = 1$.Now,for differentiability,the left-hand derivative $(LHD)$ must equal the right-hand derivative $(RHD)$ at $x = 0$.$LHD = \lim_{x 	o 0^-} f'(x) = \lim_{x 	o 0^-} (e^x + a) = e^0 + a = 1 + a$.$RHD = \lim_{x 	o 0^+} f'(x) = \lim_{x 	o 0^+} 2b(x - 1) = 2b(0 - 1) = -2b$.Equating $LHD$ and $RHD$: $1 + a = -2b$.Substituting $b = 1$: $1 + a = -2(1) \Rightarrow 1 + a = -2 \Rightarrow a = -3$.Thus,$a = -3$ and $b = 1$.

The function $f(x) = \begin{cases} e^x + ax, & x < 0 \\ b(x - 1)^2, & x \geq 0 \end{cases}$ is differentiable at $x = 0$. Then

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