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If $\mathrm{f}(\mathrm{x})=-|\mathrm{x}|$, then $\left(f^{\prime}\right) f(0)(\mathrm{x})+(f \circ f \circ f)(-\mathrm{x})=$
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The correct answer is:
$2 f(\mathrm{x})$
$f(x)=-|x|$
$\begin{aligned}(f \circ f \circ f)(x) & =f \circ f(-|x|) \\ & =f \circ(-|-| x \mid)=-|-|-|x||| \\ & =\left\{\begin{array}{cc}-x, & x \geq 0 \\ x, & x < 0\end{array}\right. \\ \text { fofof }(-x) & =f \circ f(-|-x|)\end{aligned}$
$\begin{gathered}\quad=f(-|-|-x \|)=-|-|-|-x| \mid \\ = \begin{cases}-x, & x \geq 0 \\ x, & x < 0\end{cases} \\ \Rightarrow \quad(\text { fofof })(x)+(\text { fofof })(-x)=\left\{\begin{array}{cl}-2 x, & x \geq 0 \\ 2 x & x < 0\end{array}\right. \\ \quad f(x)=\left\{\begin{array}{cl}-x, & x \geq 0 \\ x & x < 0\end{array}\right. \\ \therefore \quad(\text { fofof })(x)+(\text { fofof })(-x)=2 f(x) .\end{gathered}$
$\begin{aligned}(f \circ f \circ f)(x) & =f \circ f(-|x|) \\ & =f \circ(-|-| x \mid)=-|-|-|x||| \\ & =\left\{\begin{array}{cc}-x, & x \geq 0 \\ x, & x < 0\end{array}\right. \\ \text { fofof }(-x) & =f \circ f(-|-x|)\end{aligned}$
$\begin{gathered}\quad=f(-|-|-x \|)=-|-|-|-x| \mid \\ = \begin{cases}-x, & x \geq 0 \\ x, & x < 0\end{cases} \\ \Rightarrow \quad(\text { fofof })(x)+(\text { fofof })(-x)=\left\{\begin{array}{cl}-2 x, & x \geq 0 \\ 2 x & x < 0\end{array}\right. \\ \quad f(x)=\left\{\begin{array}{cl}-x, & x \geq 0 \\ x & x < 0\end{array}\right. \\ \therefore \quad(\text { fofof })(x)+(\text { fofof })(-x)=2 f(x) .\end{gathered}$
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