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(A) Given the function,$f(x) = \begin{cases} x^2 + bx + c, & x < 1 \ x, & x \geq 1 \end{cases}$.For $f(x)$ to be differentiable at $x = 1$,it must be

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$-2$

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(A) Given the function,$f(x) = \begin{cases} x^2 + bx + c, & x For $f(x)$ to be differentiable at $x = 1$,it must be continuous at $x = 1$ and the left-hand derivative must equal the right-hand derivative.First,for continuity at $x = 1$:$\lim_{x 	o 1^-} f(x) = f(1)$$\lim_{x 	o 1^-} (x^2 + bx + c) = 1$$1 + b + c = 1 \Rightarrow b + c = 0$ (Equation $1$).Next,for differentiability at $x = 1$,the derivatives must match:$f'(x) = \begin{cases} 2x + b, & x  1 \end{cases}$$\lim_{x 	o 1^-} f'(x) = \lim_{x 	o 1^+} f'(x)$$2(1) + b = 1 \Rightarrow 2 + b = 1 \Rightarrow b = -1$.Substituting $b = -1$ into Equation $1$:$-1 + c = 0 \Rightarrow c = 1$.Therefore,$b - c = -1 - 1 = -2$.

Define $f(x) = \begin{cases} x^2 + bx + c, & x < 1 \\ x, & x \geq 1 \end{cases}$. If $f(x)$ is differentiable at $x = 1$,then $(b - c)$ is equal to

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