Q. 79 - 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 $g.$

$f\left(x\right)=-x^2+5x; g\left(x\right)=x^2+3x-4$

Step-by-Step Solution

Answer

(a) The graph of $f$ and $g$ is 

(b) $x=2,1.$

(c) The point of intersection is $\left(2,6\right),\left(-1,6\right).$ 

(d) The shaded region is 

Part (a) Step 1. Given information

It is given that $f\left(x\right)=-x^2+5x; g\left(x\right)=x^2+3x-4.$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 $f\left(x\right)=-x^2+5x; g\left(x\right)=x^2+3x-4.$ We need to solve $f\left(x\right)=g\left(x\right).$

Step 2. Solving

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

$-x^2+5x=x^2+3x-4.$

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

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

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

   $=\frac{1\pm \sqrt{1+8}}{4}.$

   $=\frac{1\pm \sqrt{9}}{4}.$

   $=\frac{1\pm 3}{4}.$

$x=\frac{1+3}{2}=2, and x=\frac{1-3}{2}=-1.$

The values for which $f$ and $g$ are equal, when $x=2,-1.$

Part (c) Step 1. Given information

It is given that $f\left(x\right)=-x^2+5x; g\left(x\right)=x^2+3x-4.$ We need to use the result of part (b) to label the point of intersection.

Step 2. Simplify

Let $y=-x^2+5x=f\left(x\right).$

$y=g\left(x\right)=x^2+3x-4.$

When $x=2, y=-2^2+5·2=6.$

When $x=-1, y=-{\left(-1\right)}^2+5·\left(-1\right)=-6.$

The point of intersection is $\left(2,6\right),\left(-1,6\right).$

Part (d) Step 1. Given information

It is given that $f\left(x\right)=-x^2+5x; g\left(x\right)=x^2+3x-4.$ e need to draw the shaded region for $f\left(x\right)>g\left(x\right).$

Step 2. Draw the shaded region