91Ó°ÊÓ

$$\begin{gathered} f\left(x\right)=-x^2-x+1 \\ g\left(x\right)=-x^2+x+6 \end{gathered}$$

Substitute $f\left(x\right)=0$in the equation and simplify

$$\begin{gathered} f\left(x\right)=-x^2-x+1 \\ 0=-x^2-x+1 \\ {x}_1=\frac{-1+\sqrt{5}}{2},x_2=\frac{-1-\sqrt{5}}{2} \end{gathered}$$

Substitute $g\left(x\right)=0$in the given function and evaluate

$$\begin{gathered} g\left(x\right)=-x^2+x+6 \\ 0=-x^2+x+6 \\ {x}_1=-2,x_2=3 \end{gathered}$$

Equate the given function and evalute

$$\begin{gathered} f\left(x\right)=g\left(x\right) \\ -x^2-x+1=-x^2+x+6 \\ -2x=5 \\ x=-\frac{5}{2} \end{gathered}$$

Solve $f\left(x\right)>0$for the given function

First, solve for $f\left(x\right)=0$

the result is $x_1=\frac{-1+\sqrt{5}}{2},x_2=\frac{-1-\sqrt{5}}{2}$

Graph the function $f\left(x\right)=-x^2-x+1$

We have to find where the graph is above the x-axis,

So the solution is

Solve $g\left(x\right)≤0$for the given function

$g\left(x\right)=-x^2+x+6$

First, solve for $g\left(x\right)=0$and graph the function $g\left(x\right)$

$$\begin{gathered} g\left(x\right)=-x^2+x+6 \\ 0=-x^2+x+6 \\ {x}_1=-2,x_2=3 \end{gathered}$$

The graph of the function $g\left(x\right)$is

We have to find where the graph is under or on the x-axis,

So the solution is$x∈<-∞,-2]∪[3,+∞>$

To solve, $f\left(x\right)>g\left(x\right)$, sketch the function f and g in the same plane and notice where the graph of function f is above the graph of the function g.

It is shown that $f\left(x\right)=g\left(x\right)$is true for $x=-\frac{5}{2}$

For example, $$\begin{gathered} f\left(x\right)=-x^2-x+1 \\ f\left(-\frac{5}{2}\right)=\frac{21}{4} \end{gathered}$$

The point of intersection of the two graphs is localid="1650153561435" $\left(-\frac{5}{2},\frac{21}{4}\right)$

The graph is

When then the graph of the function f is above the graph of the function g,$f\left(x\right)>g\left(x\right)$

The solution for the function $f\left(x\right)â‰\yen 1$is given by

First, solve for $f\left(x\right)=1$

$$\begin{gathered} f\left(x\right)=-x^2-x+1 \\ 1=-x^2-x+1 \\ {x}_1=0,x_2=-1 \end{gathered}$$

Graph $h\left(x\right)=1$and f in the same plane and check where is the graph of the function f above the graph of the function h.

The solution is$x∈[-1,0]$