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Let $f(x)=\left\{\begin{array}{cl}-1, & -2 \leq x < 0 \\ x^{2}-1, & 0 \leq x \leq 2\end{array}\right.$ and
$g(x)=|\eta(x)|+f(x \mid) .$ Then, in the interval $(-2,2), g$ is:
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$g(x)=|\eta(x)|+f(x \mid) .$ Then, in the interval $(-2,2), g$ is:
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The correct answer is:
not differentiable at one point
$f(x)=\left\{\begin{array}{cc}-1, & -2 \leq x < 0 \\ x^{2}-1, & 0 \leq x \leq 2\end{array}\right.$
$\text { Then, } f(x \mid)=\left\{\begin{array}{cc}
-1, & -2 \leq|x| < 0 \\
|x|^{2}-1, & 0 \leq|x| \leq 2
\end{array}\right.$
$\Rightarrow f(|x|)=x^{2}-1,-2 \leq x \leq 2$
$\Rightarrow g(x)=\left\{\begin{aligned} 1+x^{2}-1, &-2 \leq x < 0 \\\left(x^{2}-1\right)+\left|x^{2}-1\right|, & 0 \leq x \leq 2 \end{aligned}\right.$
$=\left\{\begin{array}{cc}x^{2}, & -2 \leq x < 0 \\ 0, & 0 \leq x < 1 \\ 2\left(x^{2}-1\right), & 1 \leq x \leq 2\end{array}\right.$
$g^{\prime}\left(0^{-}\right)=0, g^{\prime}\left(0^{*}\right)=0, g^{\prime}\left(1^{-}\right)=0, g^{\prime}\left(1^{+}\right)=4$
$\Rightarrow g(x)$ is non-differentiable at $x=1$ $\Rightarrow g(x)$ is not differentiable at one point.
$\text { Then, } f(x \mid)=\left\{\begin{array}{cc}
-1, & -2 \leq|x| < 0 \\
|x|^{2}-1, & 0 \leq|x| \leq 2
\end{array}\right.$
$\Rightarrow f(|x|)=x^{2}-1,-2 \leq x \leq 2$
$\Rightarrow g(x)=\left\{\begin{aligned} 1+x^{2}-1, &-2 \leq x < 0 \\\left(x^{2}-1\right)+\left|x^{2}-1\right|, & 0 \leq x \leq 2 \end{aligned}\right.$
$=\left\{\begin{array}{cc}x^{2}, & -2 \leq x < 0 \\ 0, & 0 \leq x < 1 \\ 2\left(x^{2}-1\right), & 1 \leq x \leq 2\end{array}\right.$
$g^{\prime}\left(0^{-}\right)=0, g^{\prime}\left(0^{*}\right)=0, g^{\prime}\left(1^{-}\right)=0, g^{\prime}\left(1^{+}\right)=4$
$\Rightarrow g(x)$ is non-differentiable at $x=1$ $\Rightarrow g(x)$ is not differentiable at one point.
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