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Let $f(x)=a_0+a_1|x|+a_2|x|^2+a_3|x|^3$, where $a_0, a_1, a_2, a_3$ are real constants. Then $f(x)$ is differentiable at $x=0$
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only if $a_1=0$
$f^{\prime}\left(0^{+}\right)=\lim _{x \rightarrow 0}\left(a_1+2 a_1 x+3 a_3 x^2\right)=a_1$
$f^{\prime}\left(0^{-}\right)=\lim _{x \rightarrow 0}\left(-a_1+2 a_1 x-3 a_3 x^2\right)=-a_1$
$f^{\prime}\left(0^{+}\right)=f^{\prime}\left(0^{-}\right) \quad \Rightarrow a_1=0$
$f^{\prime}\left(0^{-}\right)=\lim _{x \rightarrow 0}\left(-a_1+2 a_1 x-3 a_3 x^2\right)=-a_1$
$f^{\prime}\left(0^{+}\right)=f^{\prime}\left(0^{-}\right) \quad \Rightarrow a_1=0$
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