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| The function $f(x)=|x|+||x-1|+|x-2|$ is |
Differentiable everywhere. Differentiable Nowhere. Differentiable at everywhere except 0 Differentiable everywhere except at 0,1,2 |
| Differentiable everywhere except at 0,1,2 |
| This function can be rewritten as $f(x)=\begin{cases} x+x-1+x-2=3x-3 & \text{if}{\hspace .2cm} x\geq 2\\ x+(x-1)-(x-2)=x+1 & \text{if}{\hspace .2 cm} 1\leq x \leq 2\\ x+(1-x)+(2-x)=3-x & \text{if}{\hspace .2 cm} 0\leq x \leq 1\\ -x+1-x+2-x=3-3x & \text{if}{\hspace .2 cm} x<0 \end{cases}$ Now $\lim_{x\to 2+}f(x)-f(2)/x-2=\lim_{x \to 2}3x-6/x-2=3$ but $\lim_{x\to 2+}f(x)-f(2)/x-2=\lim_{x \to 2}x+1-3/x-2=1$. So $f'(2)$ does not exist. Similarly $f'(1)$ and $f'(0)$ does not exist. |
