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(C) function $f(x)$ is differentiable at $x=a$ if the left-hand derivative $LHD = \lim_{h 	o 0^-} \frac{f(a+h)-f(a)}{h}$ and right-hand derivative $R

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$f_3(x)=\begin{cases} x^2+7x-7, & x \leq 1 \ \frac{3x-1}{2}, & x > 1 \end{cases}$

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(C) function $f(x)$ is differentiable at $x=a$ if the left-hand derivative $LHD = \lim_{h 	o 0^-} \frac{f(a+h)-f(a)}{h}$ and right-hand derivative $RHD = \lim_{h 	o 0^+} \frac{f(a+h)-f(a)}{h}$ exist and are equal.For $f_1(x)=|x|$, at $x=1$, $f_1(x)=x$. Thus $f_1'(1)=1$. It is differentiable.For $f_2(x)$, at $x=1$, $LHD = \frac{d}{dx}(1+\sin(x-1))|_{x=1} = \cos(0) = 1$. $RHD = \frac{d}{dx}(x)|_{x=1} = 1$. Since $LHD=RHD$, it is differentiable.For $f_3(x)$, at $x=1$, $LHD = \frac{d}{dx}(x^2+7x-7)|_{x=1} = 2(1)+7 = 9$. $RHD = \frac{d}{dx}(\frac{3x-1}{2})|_{x=1} = \frac{3}{2} = 1.5$. Since $LHD 
eq RHD$, $f_3(x)$ is not differentiable at $x=1$.For $f_4(x)$, at $x=1$, $f_4(x) = |x-1|+|x-2|$. For $x \leq 1$, $f_4(x) = -(x-1)-(x-2) = -2x+3$. $LHD = -2$. For $x > 1$, $f_4(x) = 1+x-x^3$. $RHD = \frac{d}{dx}(1+x-x^3)|_{x=1} = 1-3(1)^2 = -2$. Since $LHD=RHD$, it is differentiable.

The function that is not differentiable at $x=1$ is

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