If f(x) = begin{cases} e^x + ax, & x

Question

(B) Given that $f(x)$ is differentiable at $x = 0$. Since every differentiable function is continuous,$f(x)$ must be continuous at $x = 0$. Thus,$\lim

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$(-3, 1)$

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(B) Given that $f(x)$ is differentiable at $x = 0$. Since every differentiable function is continuous,$f(x)$ must be continuous at $x = 0$. Thus,$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$. $\lim_{x 	o 0^-} (e^x + ax) = e^0 + a(0) = 1$. $\lim_{x 	o 0^+} b(x - 1)^2 = b(0 - 1)^2 = b$. Equating these,we get $b = 1$. Since $f(x)$ is differentiable at $x = 0$,the left-hand derivative $(LHD)$ must equal the right-hand derivative $(RHD)$ at $x = 0$. $LHD = \frac{d}{dx}(e^x + ax) = e^x + a$. At $x = 0$,$LHD = e^0 + a = 1 + a$. $RHD = \frac{d}{dx}(b(x - 1)^2) = 2b(x - 1)$. At $x = 0$,$RHD = 2b(0 - 1) = -2b$. Setting $LHD = RHD$,we have $1 + a = -2b$. Substituting $b = 1$,we get $1 + a = -2(1) \implies 1 + a = -2 \implies a = -3$. Therefore,$(a, b) = (-3, 1)$.

If $f(x) = \begin{cases} e^x + ax, & x < 0 \\ b(x - 1)^2, & x \ge 0 \end{cases}$ is differentiable at $x = 0$,then $(a, b)$ is

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