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(C) Given $f(x) = a_0 + a_1|x| + a_2|x|^2 + a_3|x|^3$. Since $|x|^2 = x^2$ and $|x|^3 = |x| \cdot x^2$,we can write $f(x) = a_0 + a_1|x| + a_2x^2 + a_

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only if $a_1=0$

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(C) Given $f(x) = a_0 + a_1|x| + a_2|x|^2 + a_3|x|^3$. Since $|x|^2 = x^2$ and $|x|^3 = |x| \cdot x^2$,we can write $f(x) = a_0 + a_1|x| + a_2x^2 + a_3|x|x^2$. For $f(x)$ to be differentiable at $x=0$,the left-hand derivative and right-hand derivative must be equal. Right-hand derivative at $x=0$: $f'(0^+) = \lim_{h 	o 0^+} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{a_0 + a_1h + a_2h^2 + a_3h^3 - a_0}{h} = \lim_{h 	o 0^+} (a_1 + a_2h + a_3h^2) = a_1$. Left-hand derivative at $x=0$: $f'(0^-) = \lim_{h 	o 0^-} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0^-} \frac{a_0 + a_1(-h) + a_2h^2 + a_3(-h^3) - a_0}{h} = \lim_{h 	o 0^-} (-a_1 + a_2h - a_3h^2) = -a_1$. For differentiability,$f'(0^+) = f'(0^-) \implies a_1 = -a_1 \implies 2a_1 = 0 \implies a_1 = 0$. Thus,$f(x)$ is differentiable at $x=0$ only if $a_1=0$.

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 $x=0$ if and only if:

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