Q. 77 - 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 $g.$

$f\left(x\right)=-x^2+4; g\left(x\right)=-2x+1.$

Step-by-Step Solution

Answer

(a) The graph $f$of  and $g$

(b) $x=3, x=-1.$

(c) The point of intersection is $\left(3,-5\right),\left(-1,3\right).$

(d) The shaded region is,

Part (a) Step 1. Given information

It is given that $f\left(x\right)=-x^2+4; g\left(x\right)=-2x+1.$We need to graph $f$ and $g$ on the same cartesian plane.

Step 2. Graph f x and g x

Part (b) Step 1. Given information

It is given that $f\left(x\right)=-x^2+4; g\left(x\right)=-2x+1.$ We need to solve $f\left(x\right)=g\left(x\right).$

Step 2. Solving

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

$-x^2+4 =-2x+1.$

$x^2-2x-3=0.$

$x=\frac{+2\pm \sqrt{{\left(-2\right)}^2-4·1·\left(-3\right)}}{2·1}.$

  $=\frac{+2\pm \sqrt{4+12}}{2}.$

   $=\frac{+2\pm \sqrt{16}}{2}.$

   $=\frac{2\pm 4}{2}.$

$x=\frac{2+4}{2}=3,$

and $x=\frac{2-4}{2}=-1.$

The value of the function $f$ and $g$ are equal when $x=3,-1.$

Part (c) Step 1. Given information

It is given that $f\left(x\right)=-x^2+4; g\left(x\right)=-2x+1.$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 $x=3, y=-2·\left(3\right)+1=5.$

When $x=-1, y=-2·\left(-1\right)+1=3.$

The point of intersection is $\left(3,-5\right),\left(-1,3\right).$

Part (d) Step 1. Given information

It is given that $f\left(x\right)=-x^2+4; g\left(x\right)=-2x+1.$ We need to draw the shaded region for which $f\left(x\right)>g\left(x\right).$

Step 2. Draw the shaded region