Q 32 - Precalculus Enhanced with Graphing Utilities

Question

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

(a) Solve $f\left(x\right)=0$. (b) Solve $g\left(x\right)=0$. (c) Solve $f\left(x\right)=g\left(x\right)$. (d) Solve $f\left(x\right)>0$.

(e) Solve $g\left(x\right)\le 0$. (f) Solve $f\left(x\right)>g\left(x\right)$. (g) Solve $f\left(x\right)\ge 1$ .

$$\begin{gathered} f\left(x\right)=-x^2-x+1 \\ g\left(x\right)=-x^2+x+6 \end{gathered}$$

Step-by-Step Solution

Answer

(a) $x_1=\frac{-1+\sqrt{5}}{2}, x_2=\frac{-1-\sqrt{5}}{2}$

(b) $x_1=-2, x_2=3$

(c) $x=-\frac{5}{2}$

(d) The solution is 

(e) $x\in <-\infty ,-2]\cup [3,+\infty >$

(f) 

(g) $x\in [-1,0]$

Part (a) Step 1. Given Information

$$\begin{gathered} f\left(x\right)=-x^2-x+1 \\ g\left(x\right)=-x^2+x+6 \end{gathered}$$

Part (a) Step 2. Solve f x = 0

Substitute $f\left(x\right)=0$ in the equation and simplify

$$\begin{gathered} f\left(x\right)=-x^2-x+1 \\ 0=-x^2-x+1 \\ {x}_1=\frac{-1+\sqrt{5}}{2}, x_2=\frac{-1-\sqrt{5}}{2} \end{gathered}$$

Part (b) Step 1. Solve g x = 0

Substitute $g\left(x\right)=0$ in the given function and evaluate

$$\begin{gathered} g\left(x\right)=-x^2+x+6 \\ 0=-x^2+x+6 \\ {x}_1=-2, x_2=3 \end{gathered}$$

Part (c) Step 1. Solve f x = g x

Equate the given function and evalute

$$\begin{gathered} f\left(x\right)=g\left(x\right) \\ -x^2-x+1=-x^2+x+6 \\ -2x=5 \\ x=-\frac{5}{2} \end{gathered}$$

Part (d) Step 1. Solve f x &#62; 0

Solve $f\left(x\right)>0$ for the given function

First, solve for $f\left(x\right)=0$

the result is $x_1=\frac{-1+\sqrt{5}}{2}, x_2=\frac{-1-\sqrt{5}}{2}$

Graph the function $f\left(x\right)=-x^2-x+1$

We have to find where the graph is above the x-axis, 

So the solution is 

Part (e) Step 1. Solve g x &#8804; 0

Solve $g\left(x\right)\le 0$ for the given function

$g\left(x\right)=-x^2+x+6$

First, solve for $g\left(x\right)=0$ and graph the function $g\left(x\right)$

$$\begin{gathered} g\left(x\right)=-x^2+x+6 \\ 0=-x^2+x+6 \\ {x}_1=-2, x_2=3 \end{gathered}$$

 The graph of the function $g\left(x\right)$ is

We have to find where the graph is under or on the x-axis,

So the solution is $x\in <-\infty ,-2]\cup [3,+\infty >$

Part (f) Step 1. Solve f x &#62; g x

To solve, $f\left(x\right)>g\left(x\right)$, sketch the function f and g in the same plane and notice where the graph of function f is above the graph of the function g.

It is shown that $f\left(x\right)=g\left(x\right)$ is true for $x=-\frac{5}{2}$

For example, $$\begin{gathered} f\left(x\right)=-x^2-x+1 \\ f\left(-\frac{5}{2}\right)=\frac{21}{4} \end{gathered}$$

The point of intersection of the two graphs is $\left(-\frac{5}{2},\frac{21}{4}\right)$

The graph is 

When   then the graph of the function f is above the graph of the function g, $f\left(x\right)>g\left(x\right)$

Part (g) Step 1. Solve f x &#8805; 1

The solution for the function $f\left(x\right)\ge 1$ is given by

First, solve for $f\left(x\right)=1$

$$\begin{gathered} f\left(x\right)=-x^2-x+1 \\ 1=-x^2-x+1 \\ {x}_1=0, x_2=-1 \end{gathered}$$

Graph $h\left(x\right)=1$ and f in the same plane and check where is the graph of the function f above the graph of the function h.

The solution is $x\in [-1,0]$