If f(x) = |x| + |x - 1|, write the value of $\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)$
Answer
$\text{f}(\text{x})=|\text{x}|+|\text{x}+1|$When x > 1
$\text{f}(\text{x})=\text{x}+\text{x}-1=\text{2x}-1$
$\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=2$
When 0 < x < 1
$\text{f}(\text{x})=\text{x}-\text{x}+1=1$
$\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=0$
When x < 0
$\text{f}(\text{x})=-\text{x}-\text{x}+1=-\text{2x}+1$
$\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=-2$
$\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\begin{cases}2,&\text{x}>1\\0,&0<\text{x}<1\\-2,&\text{x}<0\end{cases}$
Write the value of $\lim_\limits{\text{x}\rightarrow\text{c}}\frac{\text{f}(\text{x})-\text{f}(\text{c})}{\text{x}-\text{c}}$
Answer
Since we know that,$\frac{\text{d}}{\text{dx}}(\text{f}(\text{x}))=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h}-\text{f}(\text{x}))}{\text{h}}$
Let,
$\text{h}=\text{x}-\text{c}$ and $\text{f}(\text{x})=\text{c}$
If $\text{h}\rightarrow0$ then $\text{x}\rightarrow\text{c}$
Therefore,
$\frac{\text{d}}{\text{dx}}(\text{c})=\lim_\limits{\text{x}\rightarrow\text{c}}\frac{\text{f}(\text{c}+\text{x}-\text{c})-\text{f}(\text{c})}{\text{x}-\text{c}}$
$=\lim_\limits{\text{x}\rightarrow\text{c}}\frac{\text{f}(\text{x})-\text{f}(\text{c})}{\text{x}-\text{c}}$
$=\text{f}'(\text{c})$