Q. 80 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 75–80, for the given functions $f$ and $g.$ 

(a) Graph $f$ and $g$ on the same Cartesian plane.

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

(c) Use the result of part (b) to label the points of intersection of the graphs of $f$ and $g.$

 (d) Shade the region for which $f\left(x\right)>g\left(x\right),$ that is, the region below $f$ and above $g$ 

$f\left(x\right)=-x^2+7x-6; g\left(x\right)=x^2+x-6$

Step-by-Step Solution

Answer

(a) The graph of $f$ and $g.$

(b) $x=0,3.$

(c) The points of intersection $\left(0,-6\right),\left(3,6\right).$

(d) The shaded region 

Part (a) Step 1. Given information

It is given that $f\left(x\right)=-x^2+7x-6; g\left(x\right)=x^2+x-6.$ We need to draw the graph of $f$ and $g$ in the same cartesian plane.

Step 2. Draw the graph

Part b Step 1. Given information

It is given that $f\left(x\right)=-x^2+7x-6; g\left(x\right)=x^2+x-6.$ We need to solve $f\left(x\right)=g\left(x\right).$

Step 2. Solving

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

$-x^2+7x-6=x^2+x-6.$

$2x^2-6x=0.$

$x\left(x-3\right)=0.$

$x=0, and x-3=0.$

                       $x=3.$

The values of the function $f$ and $g$ are equal, when $x=0,3.$

Part (c) Step 1. Given information

It is given that $f\left(x\right)=-x^2+7x-6; g\left(x\right)=x^2+x-6.$ We need to use the result of part (b) to label the point of intersection.

Step 2. Simplify

Let $y=f\left(x\right)=-x^2+7x-6.$

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

When $x=0, y=0^2+0-6=-6.$

When $x=3, y=3^2+3-6=6.$

The point of intersection is $\left(0,-6\right),\left(3,6\right).$

Part (d) Step 1. Given information

It is given that $f\left(x\right)=-x^2+7x-6;g\left(x\right)=x^2+x-6.$ We need to shade the region for which $f\left(x\right)>g\left(x\right).$

Step 2. Shade the region