Chapter 1.8, Problem 73E

a.

To determine

To graph: $y=\frac{1}{2}{x}^2-2x+1,y\le 0$

Then find the approximate value of x that satisfies each inequalities using graph.

Answer to Problem 73E

  

From the graph it can be concluded that $x=0.586$ and $x=3.414$ satisfies the given conditions

Explanation of Solution

Given:

  $y=\frac{1}{2}{x}^2-2x+1,y\le 0$

Graph:

For $y=\frac{1}{2}{x}^2-2x+1$ form a table as

  $\begin{array}{ccccccccc}x& 0& 1& -1& 2& -2& 3& 4& 5\\ y& 1& -1.5& 3.5& -1& 7& -0.5& 1& 3.5\end{array}$

Plot the points and join them

  

Now $y\le 0$ implies the region below x-axis because that region will always be negative that is $y\le 0$

Hence the required graph will be

  

From the graph it can be concluded that $x=0.586$ and $x=3.414$ satisfies the given conditions

b.

To determine

To graph: $y=\frac{1}{2}{x}^2-2x+1,y\ge 7$

Then find the approximate value of x that satisfies each inequalities using graph.

Answer to Problem 73E

  

From the graph it can be concluded that $x=-2$ and $x=6$ satisfies the given conditions

Explanation of Solution

Given:

  $y=\frac{1}{2}{x}^2-2x+1,y\ge 7$

Graph:

For $y=\frac{1}{2}{x}^2-2x+1$ form a table as

  $\begin{array}{ccccccccc}x& 0& 1& -1& 2& -2& 3& 4& 5\\ y& 1& -1.5& 3.5& -1& 7& -0.5& 1& 3.5\end{array}$

Plot the points and join them

  

Now $y\ge 7$ implies the region above the line $y=7$ including the line

Hence the required graph will be

  

From the graph it can be concluded that $x=-2$ and $x=6$ satisfies the given conditions