Question

$f\left(x\right)=\{\begin{array}{c}x;x<1\\ 3-x;1\le x\le 3\end{array}$ then $f^{{}^{\prime }}\left(x\right)=$

• A. $2$
• B. $0$
• C. $-1$
• D. Does not exist

Answer 1

Answer: (D)

Does not exist

$f\left(x\right)=x;x<1$
$f\left(x\right)=3-x;1\le x\le 3$
If $f^{\prime }(x{)}_{x\to 1^{-}}=f^{\prime }(x{)}_{x\to 1^{+}}=f^{\prime }\left(1\right)$
So, $f^{\prime }(x{)}_{x\to 1^{-}}=1$
$f^{\prime }(x{)}_{x\to 1^{+}}=-1$
$\because f^{\prime }(x{)}_{x\to 1^{-}}\ne f^{\prime }(x{)}_{x\to 1^{+}}$
So, $f^{\prime }\left(x\right)$ doesn't exist.