Q. 75 - 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 $g.$

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

Step-by-Step Solution

Answer

(a) The graph of $f\left(x\right)$ and $g\left(x\right).$

(b) $x=3, x=-1.$

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

(d) The shaded region for $f\left(x\right)>g\left(x\right).$ is 

Part (a) Step 1. Given information

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

Step 2. Draw the graph

Part (b) Step 1. Given information

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

Step 2. Simplify

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

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

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

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

   $=\frac{2\pm \sqrt{16}}{2}.$

   $=\frac{2\pm 4}{2}.$

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

$x=3.$                $x=-1.$

The values of the function $f$ and $g$ are equal when $x_1=3, x_2=-1.$

Part c Step 1. Given information

It is given that $f\left(x\right)=2x-1, g\left(x\right)=x^2-4.$ We need to use the result of part (b) to label the point of intersecting.

Step 2. Simplify

When $x_1=3, y=2·3-1\iff y=5.$

When $x_2=-1, y=2·\left(-1\right)-1\iff y=-3.$

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

Part d Step 1. Given information

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

Step 2. Shade the region