If f(x) = begin{cases} x^2 & text{if } x leqs | vedclass

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(A) For $f(x)$ to be derivable at $x = x_0$,it must be continuous at $x = x_0$. Continuity at $x = x_0$ implies $\lim_{x 	o x_0^-} f(x) = \lim_{x 	o

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$2x_0, -x_0^2$

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(A) For $f(x)$ to be derivable at $x = x_0$,it must be continuous at $x = x_0$. Continuity at $x = x_0$ implies $\lim_{x 	o x_0^-} f(x) = \lim_{x 	o x_0^+} f(x) = f(x_0)$. $\lim_{x 	o x_0^-} x^2 = x_0^2$ and $\lim_{x 	o x_0^+} (ax + b) = ax_0 + b$. So,$x_0^2 = ax_0 + b$  (Equation $1$). For differentiability at $x = x_0$,the left-hand derivative must equal the right-hand derivative. $f'(x) = \begin{cases} 2x & 	ext{if } x  x_0 \end{cases}$. Equating derivatives at $x = x_0$: $2x_0 = a$. Substitute $a = 2x_0$ into Equation $1$: $x_0^2 = (2x_0)x_0 + b$. $x_0^2 = 2x_0^2 + b \implies b = -x_0^2$. Thus,$a = 2x_0$ and $b = -x_0^2$.

If $f(x) = \begin{cases} x^2 & \text{if } x \leqslant x_0 \\ ax + b & \text{if } x > x_0 \end{cases}$ is derivable for all $x \in \mathbb{R}$,then the values of $a$ and $b$ are respectively:

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