Chapter 10, Problem 3CR

To determine

The values of $x$, for which $6-x\ge x^2$

Answer to Problem 3CR

Solution:

All the values of the interval $\left[-3,2\right]$ satisfies $6-x\ge x^2$

Explanation of Solution

Given information:

The inequality $6-x\ge x^2$

Solving $6-x\ge x^2$ is equivalent to finding an interval, where the graph of $6-x$ is above the graph of $x^2$

Write the inequality so that the polynomial expression is on the left side and zero is on the right side.

Subtracting $x^2$ from both sides,

  $\Rightarrow 6-x-x^2\ge 0$ .

To determine the real zeroes of the function $6-x-x^2$

  $6-x-x^2=0$

  $\Rightarrow x^2+x-6=0$

  $\Rightarrow \left(x+3\right)\left(x-2\right)=0$

  $\Rightarrow x=-3$ or $x=2$

The real zeroes of the function are $x=-3$ and $x=2$

Use the real zeroes to separate the real number line into three intervals as

  $(-\infty ,-3),(-3,2),(2,\infty )$ .

To select a test point from each interval and determine whether $f(x)$ is positive or negative.

  $\begin{array}{cccc}\text{Interval}& (-\infty ,-3)& (-3,2)& (2,\infty )\\ \text{Number}\text{chosen}& -4& 0& 4\\ \text{Value}\text{of}6-x-x^2& \begin{array}{l}6-\left(-4\right)-{\left(-4\right)}^2\\ =-6\end{array}& \begin{array}{l}6-0-{\left(0\right)}^2\\ =6\end{array}& \begin{array}{l}6-4-{\left(4\right)}^2\\ =-14\end{array}\\ \text{Conclusion}& \text{Negative}& \text{Positive}& \text{Negative}\end{array}$

  $6-x-x^2$ is positive for the interval $\left(-3,2\right)$

For the inequality $6-x-x^2\ge 0$, need to check at end points of the interval also.

For end points $x=-3,2$

Endpoint $x=-3$, $6-\left(-3\right)-{\left(-3\right)}^2=6+3-9=0$

Hence, $x=-3$ holds the inequality $6-x\ge x^2$ .

Endpoint $x=2$, $6-\left(2\right)-{\left(2\right)}^2=6-2-4=0$

Hence $x=2$ holds the inequality $6-x\ge x^2$

Therefore, the endpoints $x=-3$ and $x=2$ are included in the solution.

Thus, the inequality $6-x\ge x^2$ has the solution $\left[-3,2\right]$