Question

The function $f(x) = \begin{cases} 5x-4 & \quad \text{for } 0<\,x \leq \,1 \\ 4x^{2}-3x & \quad \text{for } 1<\, x<\,2, \\ 3x+4 & \quad \text{for } x \,\geq\, 2, \end{cases} $
then which of the following is not true?

• A. Continuous at $x=1$ and $x=2$
• B. Continuous at $x=1$ but not derivable at $x=2$
• C. Continuous at $x=2$ but not derivable at $x=1$
• D. None of these

Answer 1

Answer: (C)

$f(x) = \begin{cases} 5x-4 & \quad \text{for } 0<\,x \leq \,1 \\ 4x^{2}-3x & \quad \text{for } 1<\, x<\,2, \\ 3x+4 & \quad \text{for } x \,\geq\, 2, \end{cases} $
and $f'(x) = \begin{cases} 5 & \quad \text{for } 0<\,x < \,1 \\ 8x-3 & \quad \text{for } 1<\, x<\,2 \\ 3& \quad \text{for } x>\,2 \end{cases} $
$f\left(1^{-}\right)=1 ; f\left(1^{+}\right)=1 ; f(1)=1$ and $f^{\prime}\left(1^{-}\right)=5 ; f^{\prime}\left(1^{+}\right)=5$
$f(2)=f\left(2^{+}\right)=10, f\left(2^{-}\right)=10 ; f^{\prime}\left(2^{+}\right)=3 ; f\left(2^{-}\right)=13$
Thus $f(x)$ is continuous at $x=1$ and $x=2,$ differential at $x=1$ but non-differentiable at $x=2$