Question

Let $f\left(x\right)=x^4-8x^3+22x^2-24x$ and $g\left(x\right)=\{\begin{array}{c}\begin{array}{ccc}min\left(f\left(t\right)\right),& x\le t\le x+1,& -1\le x\le 1\end{array}\\ \begin{array}{cc}x-10& x>1\end{array}\end{array}$.

Then, in the interval $[-1,\infty )$, $g\left(x\right)$ is

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

Answer 1

Answer: (A), (D)

The correct options are
A Continuous for all $x$
D Not differentiable at $x=1$

Here, $f\left(x\right)=x^4-8x^3+22x^2-24x$

$\Rightarrow f^{\prime }\left(x\right)=4x^3-24x^2+44x-24$

$\Rightarrow f^{\prime }\left(x\right)=4(x-1)(x-2)(x-3)$

which shows $f\left(x\right)$is increasing in $[1,2]\cup [3,\infty )$ and decreasing in $(-\infty ,1]\cup [2,3].$

Thus, minimum $f\left(x\right);x\le t\le x+1,-1\le x\le 1$

$\Rightarrow$ minimum $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}f(x+1),& -1\le x\le 0\end{array}\\ \begin{array}{cc}f\left(1\right),& 0<x\le 1\end{array}\end{array}$

Thus, $g\left(x\right)=\{\begin{array}{c}\begin{array}{cc}f(x+1),& -1\le x\le 0\end{array}\\ \begin{array}{cc}f\left(1\right),& 0<x\le 1\end{array}\\ \begin{array}{cc}x-10,& x>1\end{array}\end{array}=\{\begin{array}{c}\begin{array}{cc}{(x+1)}^4-{8(x+1)}^3+22{(x+1)}^2-24(x+1),& -1\le x\le 0\end{array}\\ \begin{array}{cc}1-8+22-24,& 0<x\le 1\end{array}\\ \begin{array}{cc}x-10,& x>1\end{array}\end{array}$

$\Rightarrow g\left(x\right)=\{\begin{array}{c}\begin{array}{cc}{x}^4-4x^3+4x^2-9,& -1\le x\le 0\end{array}\\ \begin{array}{cc}-9,& 0<x\le 1\end{array}\\ \begin{array}{cc}x-10,& x>1\end{array}\end{array}$

Also, $g^{\prime }\left(x\right)=\{\begin{array}{c}\begin{array}{cc}4x^3-12x^2+8x,& -1\le x\le 0\end{array}\\ \begin{array}{cc}0,& 0<x\le 1\end{array}\\ \begin{array}{cc}1,& x>1\end{array}\end{array}$

which clearly shows $g\left(x\right)$ is continuous in $[-1,\infty )$, but not differentiable at $x=1$.