Chapter 3.1, Problem 32AYU

In parts (a) - (f), use the following figure.

a. Solve $g\left(x\right)=20$.

b Solve $g\left(x\right)=60$.

c. Solve $g\left(x\right)=0$.

d. Solve $g\left(x\right)>20$.

c. Solve $g\left(x\right) \le 60$.

f. Solve $0< g\left(x\right)<60$.

To determine

To calculate: Solve the following function by using the given graph:

a. Solve $g(x)=20$

b. Solve $g(x)=60$

c. Solve $g(x)=0$

d. Solve $g(x)>20$

e. Solve $g(x)\le 60$

f. Solve $0<g(x)<60$

Answer to Problem 32AYU

Solution:

a. $x=40$

b. $x=88$

c. $x=-40$

d. $x>40$

e. $x\le 88$

f. $-40<x<88$

Explanation of Solution

Given:

The given figure is

Formula Used:

We know that the coordinates of a point are always written in the form $\left(x,f(x)\right)$.

This means that a point in the graph represents the value of the function at any point $x$.

Calculation:

a. Solve $g(x)=20$

From the given figure, we can see that $(5,20)$ is the point that satisfies $g(x)=20$.

Therefore, we have $(x,20)=(5,20)$.

Thus, we get the solution as $x=5$.

b. Solve $g(x)=60$

From the given figure, we can see that $(-15,60)$ is the point that satisfies $g(x)=60$.

Therefore, we have $(x,60)=(-15,60)$.

Thus, we get the solution as $x=-15$.

c. Solve $g(x)=0$

From the given figure, we can see that $(15,0)$ is the point that satisfies $g(x)=0$.

Therefore, we have $(x,0)=(15,0)$.

Thus, we get the solution as $x=15$.

d. Solve $g(x)>20$

When $g(x)=20$, we have $x=5$.

Therefore, when $g(x)>20$, we get $x>5$.

e. Solve $g(x)\le 60$

When $g(x)=60$, we have $x=-15$.

Therefore, when $g(x)\le 60$, we get $x\le -15$.

f. Solve $0<g(x)<60$

When $g(x)=0$, we have $x=15$.

Similarly, when $g(x)=60$, we have $x=-15$.

Therefore, when $g(x)>0$, we have $x>15$.

Similarly, when $g(x)<60$, we have $x<-15$.

Therefore, on combining the above 2 results, we get that when $0<g(x)<60$, we have $15<x<-15$.