Question

Let $f\left(x\right)\{\begin{array}{c}x\forall x<1\\ 2-x\forall 1\le x\le 2\\ {-x}^2+3x-2\forall x>2\end{array}$, then $f\left(x\right)$ is

• A. differentiable at $x=1$
• B. differentiable at $x=2$
• C. differentiable at $x=1$ and $x=2$
• D. differentiable at $x=0$

Answer 1

Answer: (B), (D)

differentiable at $x=0$
differentiable at $x=2$

$f\left(x\right)=\{\begin{array}{c}x,x<1\\ 2-x,1\le x\le 2\\ -2+3x-x^2,x>2\end{array}$

$\Rightarrow f^{\prime }\left(x\right)=\{\begin{array}{c}1,x<1\\ -1,1<x<2\\ 3-2x,x>2\end{array}$
$f^{\prime }\left(2^{-}\right)=-1$

$f^{\prime }\left(2^{+}\right)=3-2\times 2=-1$

Hence, $f\left(x\right)$ is differentiable at $x=2$.

$f^{\prime }\left(1^{-}\right)=1$

$f^{\prime }\left(1^{+}\right)=-1$

Hence, $f\left(x\right)$ is not differentiable at $x=1$

For $x<1$, $f^{\prime }\left(x\right)$ is $1$. Hence, it is differentiable at $x=0$.