Question

Let $f$ be defined by $f\left(x\right)=\left|\begin{array}{c}x+2\end{array}\right|+\left|\begin{array}{c}x\end{array}\right|+\left|\begin{array}{c}x-2\end{array}\right|$ for $x\in R$ then

• A. $f^{\prime }(-2+)$ does not exist
• B. $f^{\prime }(2-)$ does not exist
• C. $f^{\prime }(2+)=3$
• D. $f^{\prime }(0+)=2$

Answer 1

The correct option is C $f^{\prime }(2+)=3$

$f\left(x\right)=|x+2|+\left|x\right|+|x-2|f(-2^{+})=x+2-x-x+2=4-x\Rightarrow f^{\prime }(-2^{+})=-1f\left(2^{+}\right)=x+2+x+x-2=3x\Rightarrow f^{\prime }\left(2^{+}\right)=3f\left(2^{-}\right)=x+2+x+2-x=x+4\Rightarrow f^{\prime }\left(2^{-}\right)=1f\left(0^{+}\right)=x+2+x+2-x=x+2\Rightarrow f^{\prime }\left(0^{+}\right)=1$