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(A) Step $1$: Find the derivative $f'(x)$.$f'(x) = \begin{cases} 2ax & 	ext{if } x  -1 \end{cases}$Step $2$: Since $f

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$a = 2, b = 3$

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(A) Step $1$: Find the derivative $f'(x)$.$f'(x) = \begin{cases} 2ax & 	ext{if } x  -1 \end{cases}$Step $2$: Since $f'(x)$ is continuous everywhere, it must be continuous at $x = -1$. Thus, $\lim_{x 	o -1^-} f'(x) = \lim_{x 	o -1^+} f'(x)$.$2a(-1) = 2b(-1) + a \implies -2a = -2b + a \implies 3a = 2b \implies b = \frac{3}{2}a$.Step $3$: For $f'(x)$ to exist at $x = -1$, $f(x)$ must be continuous at $x = -1$. Thus, $\lim_{x 	o -1^-} f(x) = f(-1)$.$a(-1)^2 + b = b(-1)^2 + a(-1) + 4 \implies a + b = b - a + 4 \implies 2a = 4 \implies a = 2$.Step $4$: Substitute $a = 2$ into $b = \frac{3}{2}a$.$b = \frac{3}{2}(2) = 3$. Therefore, $a = 2, b = 3$.

If the derivative of the function $f(x) = \begin{cases} ax^2 + b & \text{if } x < -1 \\ bx^2 + ax + 4 & \text{if } x \geq -1 \end{cases}$ is continuous everywhere, then:

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