Question

The function
$f\left(x\right)=\{\begin{array}{cc}5x-4& \text{for}0<x\le 1\\ 4x^2-3x& \text{for}1<x<2\\ 3x+4& \text{for}x\ge 2\end{array}$ is

• A. Continuous at $x=1$ and $x=2$
• B. Continuous at $x=1$ but not differentiable at $x=2$
• C. Continuous at $x=2$ but not differentiable at $x=1$
• D. Continuous at $x=1,x=2$ but not differentiable at $x=1,x=2.$

Answer 1

Answer: (A), (B)

Continuous at $x=1$ but not differentiable at $x=2$

$f\left(x\right)=\{\begin{array}{cc}5x-4& \text{for}0<x\le 1\\ 4x^2-3x& \text{for}1<x<2\\ 3x+4& \text{for}x\ge 2\end{array}$

$f\left(1^{-}\right)=1;f\left(1^{+}\right)=1;f\left(1\right)=1$
$f\left(2^{-}\right)=10;f\left(2^{+}\right)=10;f\left(2\right)=10$
Thus continuous at $x=1,2$
Now,
$f^{\prime }\left(x\right)=\{\begin{array}{cc}5& \text{for}0<x\le 1\\ 8x-3& \text{for}1<x<2\\ 3& \text{for}x\ge 2\end{array}$

$f^{\prime }\left(1^{-}\right)=5;f^{\prime }\left(1^{+}\right)=5$
​​​​​​​$f^{\prime }\left(2^{-}\right)=13;f^{\prime }\left(2^{+}\right)=3$
Thus non differentiable at $x=2$