Chapter 5, Problem 24P

To determine

To find: the maximum and minimum values of g(c) if $-2\le c\le 2$ .

Answer to Problem 24P

Minimum value $-\frac{7}{12}$ ,Maximum value $\frac{7}{3}$

Explanation of Solution

Given information: For any number c, we let $f_c(x)$ be the smaller of the two numbers

  ${(x-c)}^2\text{and}{(x-c-2)}^2$ , then define $g(c)={\displaystyle \underset{0}{\overset{1}{\int }}{f}_c}(x)dx$

Calculation:

  $f_c(x)=\min ({(x-c)}^2,{(x-c-2)}^2)$ .

  $y={(x-c)}^2$ is a parabola, opening upward , with a vertex at x=c .

  $y={(x-(c+2))}^2$ is a parabola, opening upward , with a vertex at x=c + 2.

The intersection of these two parabolas is where: ${(x-c)}^2={(x-(c+2))}^2$ .It is at x=c +1. It is given, that $(-2\le c\le 2)$ and:

  $g(c)={\displaystyle \underset{0}{\overset{1}{\int }}{f}_c}(x)dx$ .

  $(c+1<0)\Rightarrow (c<-1)andmorespecifically(-2<c<-1)$

  $\Rightarrow f_c(x)={(x-(c+2))}^2$

  $\Rightarrow g(c)={\displaystyle \underset{0}{\overset{1}{\int }}{(x-(c+2))}^2}dx$

  $\Rightarrow \left[-\frac{x^3}{3}-(c+2)x^2+{(c+2)}^{2}x\right]\begin{array}{c}1\\ 0\end{array}=c^2+3c+\frac{5}{3}$

  $(0<c+1<1)\Rightarrow (-1<c<0)$

  $\Rightarrow f_c(x)={(x-c)}^2\text{for}0<x<(c+1)$

  $\Rightarrow f_c(x)={(x-(c+2))}^2\text{for}(c+1)<x<1$

  $\Rightarrow g(c)={\displaystyle \underset{0}{\overset{(c+1)}{\int }}{(x-c)}^2}dx+{\displaystyle \underset{(c+1)}{\overset{1}{\int }}{(x-(c+2))}^2}dx$

  $\Rightarrow \left[-\frac{x^3}{3}-c{x}^2+c^{2}x\right]\begin{array}{c}(c+1)\\ 0\end{array}+\left[\frac{x^3}{3}-(c+2)x^2+{(c+2)}^{2}x\right]\begin{array}{c}1\\ (c+1)\end{array}=-c^2-c+\frac{2}{3}$

  $(1<c+1)\Rightarrow (0<c)andmorespecifically(0<c<2)$

  $\Rightarrow f_c(x)={(x-c)}^2$

  $\Rightarrow g(c)={\displaystyle \underset{0}{\overset{1}{\int }}{(x-c)}^2}dx$

  $\Rightarrow \left[\frac{x^3}{3}-c{x}^2+c^{2}x\right]\begin{array}{c}1\\ 0\end{array}=c^2-c+\frac{1}{3}$

To summarize the three cases:

  $\begin{array}{l}g(c)=c^2+3c+\frac{5}{3}\text{for}-2<c<-1\\ g(c)=-c^2-c+\frac{2}{3}\text{for}-1<c<0\\ g(c)=c^2-c+\frac{1}{3}\text{for}0<c<2\end{array}$

To find the absolute maximum and minimum, graph this piecewise function and sec:

Minimum ( $-\frac{3}{2},-\frac{7}{12}$ ) and Maximum (2, $\frac{7}{3}$ ).

Minimum value $-\frac{7}{12}$ ,Maximum value $\frac{7}{3}$ .