Chapter 3.5, Problem 28AYU

In Problems 25-32, use the given functions $f$ and $g$.

(a) $Solvef\left(x\right)=0$.

(b) $Solveg\left(x\right)=0$.

(c) $Solvef\left(x\right)=g\left(x\right)$.

(d) $Solvef\left(x\right)>0$.

(e) $Solveg\left(x\right)\le 0$.

(f) $Solvef\left(x\right)\text{ > }g\left(x\right)$.

(g) $Solvef\left(x\right)\ge \text{ 1}$.

$f\left(x\right)=-x^2\text{ + 4}$

$g\left(x\right)=-x-\text{ 2}$

To determine

To find:

a. The solution of $f\left(x\right)=0$.

b. The solution of $g\left(x\right)=0$.

c. The solution of $f\left(x\right)=g\left(x\right)$.

d. The solution of $f\left(x\right)>0$.

e. The solution of $g\left(x\right)\le 0$.

f. The solution of $f\left(x\right)>g\left(x\right)$.

g. The solution of $f\left(x\right)\ge 1$.

Answer to Problem 28AYU

Solution:

a. $x=2,-2$

b. $x=-2$

c. $x=3,-2$

d. $-2<x<2$

e. $x\ge -2$

f. $-2<x<3$

g. $-\sqrt{3}\le x\le \sqrt{3}$

Explanation of Solution

Given:

The functions are:

$f\left(x\right)=-x^2+4$

$g\left(x\right)=-x-2$

Calculation:

a. $f\left(x\right)=0$

$-x^2+4=0$

$x^2-4=0$

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

$x=\pm 2$

Thus the solutions of $f\left(x\right)=0$ are $x=2,-2$.

b. $g\left(x\right)=0$

$-x-2=0$

$x+2=0$

$x=-2$

Thus the solution of $g\left(x\right)=0$ is $x=-2$.

c. $f\left(x\right)=g\left(x\right)$

$-x^2+4=-x-2$

$x^2-x-2-4=0$

$x^2-x-6=0$

$x^2-3x+2x-6=0$

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

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

$x=3,-2$

Thus the solutions of $f\left(x\right)=g\left(x\right)$ are $x=3,-2$.

d. $f\left(x\right)>0$

$-x^2+4>0$

$x^2-4<0$

$\left(x+2\right)\left(x-2\right)<0$

$\left(x+2\right)>0$ and $\left(x-2\right)<0$

$x>-2$ and $x<2$

$-2<x<2$

Thus the solutions of $f\left(x\right)>0$ are $-2<x<2$.

e. $g\left(x\right)\le 0$

$-x-2\le 0$

$x+2\ge 0$

$x\ge -2$

Thus the solutions of $g\left(x\right)\le 0$ are $x\ge -2$.

f. $f\left(x\right)>g\left(x\right)$

$-x^2+4>-x-2$

$x^2-x-2-4<0$

$x^2-x-6<0$

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

$\left(x-3\right)<0$ and $\left(x+2\right)>0$

$x<3$ or $x>-2$

Thus the solutions of $f\left(x\right)>g\left(x\right)$ are $-2<x<3$.

g. $f\left(x\right)\ge 1$

$-x^2+4\ge 1$

$x^2+1-4\le 0$

$x^2-3\le 0$

$x^2\le 3$

$-\sqrt{3}\le x\le \sqrt{3}$

Thus the solutions of $f\left(x\right)\ge 1$ is $-\sqrt{3}\le x\le \sqrt{3}$.