Chapter 3.3, Problem 67AYU

In Problems 77-82, for the given functions $f\text{ and }g$,

(a) Graph $f\text{ and }g$ 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 $f\text{ and }g$.

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

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

To determine

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

b. Solve $f(x)=g(x)$.

c. Label the points of intersection.

d. Shade the region below $f$ and above $g$.

Answer to Problem 67AYU

a.

b. The point of intersection is $(2,6)$ and $(-1,-6)$.

c.

d.

Explanation of Solution

Given:

The functions are

$f(x)=-x^2+5x$

$g(x)=x^2+3x - 4$

Calculation:

a. The graph of the given functions are

b. Now, we have to solve for $f(x)=g(x)$. Therefore, we have

$x^2+3x-4={\mathrm{- x}}^2+5x$

$\Rightarrow 2x^2-2x-4=0$

$\Rightarrow x^2-x-2=0$

$\Rightarrow (x-2)<mtext> </mtext>(x+1)=0$

This implies that

$x-2=0$   $x+1=0$

and

$\Rightarrow x=2$   $x=-1$

At $x=2$, we have $f(x)=-{(2)}^2+5(2)=6$

At $x=-1$, we have $f(x)=-{(-1)}^2+5(-1)=-6$.

Thus, we get the point of intersection as $(2,6)$ and $(-1,-6)$.

c. The label the point of intersection is shown below:

d. The shaded region below $f$ and above $g$ is shown below: