Question

Let $f\left(x\right)=x^3-x^2+x+1$ and $g\left(x\right)=\{\begin{array}{c}\begin{array}{ccc}max\left\{f\left(t\right)\right\},& 0\le t\le x& 0\le x\le 1\end{array}\\ \begin{array}{cc}3-x,& 1<x\le 2\end{array}\end{array}$.
Then in the interval $[0,2]$, $g\left(x\right)$ is

• A. Continuous for all $x$
• B. Differentiable for all $x$
• C. Discontinuous at $x=1$
• D. Not differentiable at $x=1$

Answer 1

Answer: (A), (D)

Continuous for all $x$
Not differentiable at $x=1$

Here, $f\left(x\right)=x^3-x^2+x+1$

$\Rightarrow f^{\prime }\left(x\right)=3x^2-2x+1,$ which is strictly increasing in $(0,2)$

$\therefore g\left(x\right)=\{\begin{array}{c}\begin{array}{cc}f\left(x\right);& 0\le x\le 1\end{array}\\ \begin{array}{cc}3-x;& 1<x\le 2\end{array}\end{array}$

$[$As $f\left(x\right)$ is increasing so $f\left(x\right)$ is maximum when $0\le t\le x]$

So, $g\left(x\right)=\{\begin{array}{c}\begin{array}{cc}{x}^3-x^2+x+1;& 0\le x\le 1\end{array}\\ \begin{array}{cc}3-x;& 1<x\le 2\end{array}\end{array}$

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

which clearly shows $g\left(x\right)$ is continuous for all $x\in [0,2]$, but $g\left(x\right)$ is not differentiable at $x=1$