Question

Let $f\left(x\right)=\{\begin{array}{cc}{x}^2/2& if0\le x\le 1\\ 2x^2-3x+3/2& if1\le x\le 2\end{array}$ then

• A. $f^{\prime }$ is not a continuous function
• B. $f^{\prime \prime }$ is not continuous at $x=1$
• C. $f$ is not differentiable at $x=1$
• D. $f$ is not continuous at $x=1$

Answer 1

Answer: (B)

$f^{\prime \prime }$ is not continuous at $x=1$

We have $\underset{x\to 1+}{lim}f\left(x\right)=\frac{1}{2}$ and $\underset{x\to 1-}{lim}f\left(x\right)=\frac{1}{2}$

So that $f$ is a continuous function. Also

$f^{\prime }\left(x\right)=\{\begin{array}{c}\begin{array}{cc}x,& 0\le x<1\end{array}\\ \begin{array}{cc}4x-3,& 1\le x<2\end{array}\end{array}$

$\Rightarrow f^{\prime }(1+)=\underset{h\to 0+}{lim}\frac{2{(1+h)}^2-3(1+4)+\frac{3}{2}-\frac{1}{2}}{h}$

$=\underset{h\to 0+}{lim}\frac{{2h}^2+h}{h}=1$
Similarly $f^{\prime }(1-)=1$.

hence $f^{\prime }$ is a continuous function. Also

$f^{\prime \prime }\left(x\right)=\{\begin{array}{c}\begin{array}{cc}1.& 0\le x<1\end{array}\\ \begin{array}{cc}4,& 1\le x<2\end{array}\end{array}$

$\Rightarrow f^{\prime \prime }(1+)={lim}_{h\to 0+}\frac{f^{\prime }(1+h)-f^{\prime }\left(1\right)}{h}$
$=\underset{h\to 0+}{lim}\frac{4(1+h)-3-1}{h}=4$
Similarly $f^{\prime \prime }(1-)=1$
Hence $f"$ is not defined at $x=1$,
so it is not continuous at $x=1$