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(D) For $-2 < x < 0$,$f(x) = \max(4 - x^2, 1 + x^2)$.$4 - x^2 = 1 + x^2 \implies 2x^2 = 3 \implies x = -\sqrt{3/2}$.Thus,$f(x) = 1 + x^2$ for $-2 < x

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$(B)$ or $(C)$ both

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(D) For $-2 $4 - x^2 = 1 + x^2 \implies 2x^2 = 3 \implies x = -\sqrt{3/2}$.Thus,$f(x) = 1 + x^2$ for $-2 For $0 $4 - x^2 = 1 + x^2 \implies x = \sqrt{3/2}$.Thus,$f(x) = 1 + x^2$ for $0 At $x = 0$,$\lim_{x 	o 0^-} f(x) = \max(4, 1) = 4$ and $\lim_{x 	o 0^+} f(x) = \min(4, 1) = 1$.Since $\lim_{x 	o 0^-} f(x) 
eq \lim_{x 	o 0^+} f(x)$,$f(x)$ is discontinuous at $x = 0$.Also,$f(x)$ is non-differentiable at $x = -\sqrt{3/2}$,$x = 0$,and $x = \sqrt{3/2}$.Therefore,$f(x)$ has a point of discontinuity and is not differentiable at more than one point. Hence,both $(B)$ and $(C)$ are correct.

Let $f(x)$ be defined in $[-2, 2]$ by
$f(x) = \begin{cases} \max(4 - x^2, 1 + x^2), & -2 < x < 0 \\ \min(4 - x^2, 1 + x^2), & 0 < x < 2 \end{cases}$
Then $f(x)$:

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Let $f(x)$ be defined in $[-2, 2]$ by$f(x) = \begin{cases} \max(4 - x^2, 1 + x^2), & -2 < x < 0 \\ \min(4 - x^2, 1 + x^2), & 0 < x < 2 \end{cases}$Then $f(x)$: