Q. 78 - 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 $g.$

$f\left(x\right)=-x^2+9; g\left(x\right)=2x+1.$

 

Step-by-Step Solution

Answer

(a) The graph of  $f$and $g$ is 

(b) $x=2,-4.$

(c) The point of intersection is $\left(2,5\right),\left(-4,-7\right).$

(d) The shaded region 

Part (a) Step 1. Given information

It is given that $f\left(x\right)=-x^2+9; g\left(x\right)=2x+1.$ We need to draw $f$ and $g$ in the same cartesian plane.

Step 2. Draw the graph

Part (b) Step 1. Given information

It is given that $f\left(x\right)=-x^2+9; g\left(x\right)=2x+1.$ We need to solve $f\left(x\right)=g\left(x\right).$

Step 2. Solving

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

$-x^2+9=2x+1.$

$x^2+2x-8=0.$

$x=\frac{-2\pm \sqrt{2^2-4·1·\left(-8\right)}}{2·1}.$

    $=\frac{-2\pm \sqrt{4+32}}{2}.$

    $=\frac{-2\pm 6}{2}.$

$x=\frac{-2+6}{2}=2, x=\frac{-2-6}{2}=-4.$

The values for the function $f$ and $g$ are equal when $x_1=2,-4.$

Part (c) Step 1. Given information

It is given that $f\left(x\right)=-x^2+9; g\left(x\right)=2x+1.$We need to use the result of part (b) to label the point of intersection.

Step 2. Simplify

Let $y=f\left(x\right)=-x^2+9.$

$y=g\left(x\right)=2x+1.$

When $x=2, y=2·2+1=5.$

When $x=-4, y=2·\left(-4\right)+1=-7.$

The point of intersection is $\left(2,5\right),\left(-4,-7\right).$

Part (d) Step 1. Given information

It is given that $f\left(x\right)=-x^2+9; g\left(x\right)=2x+1.$ We need to draw the shaded region for $f\left(x\right)>g\left(x\right).$

Step 2. Draw the shaded region