Question

Given the function. find the values of x that make the function less than or equal to zero. Write the solution in interval notation.

Answer 1

Ok, so

We got the function:

$R(x)=\frac{x-8}{x+2}$

We want to find the values which make R(x) less than or equal to zero.

$\begin{gathered} R(x)\leq0\text{ } \\ \end{gathered}$

So, we got the following inequation:

$\frac{x-8}{x+2}\leq0$

And we're going to solve it:

First of all, we're going to equal the numerator and the denominator to zero and then solve for x.

This is,

$\begin{gathered} x-8=0 \\ x=8 \end{gathered}$$\begin{gathered} x+2=0 \\ x=-2 \end{gathered}$

Those numbers that we just found, are called critical points.

Now, re're going to draw three lines, and we're going to sit both points in each line. Each line represents the numerator, the denominator, and the entire fraction.

Now, we're going to assign a sign (positive or negative) for each interval, according to the sign that the equation could take.

For example, if we analyze the first line, we notice that if we take greater values than 8, the expression x-8 will be positive. But if we take smaller values than 8, the expression will be negative:

So, we do the same thing with the second line.

if we take greater values than -2, the expression will be positive.

If we take smaller values than -2, the expression will be negative.

We represent this:

Finally, we multiply the signs:

We can see that we want R(x) less than or equal to zero, so we should take the negative interval between -2 and 8.

Now, we just notice that the denominator of the function can't be -2, because the division between 0 is not possible. So -2 is not a solution.

Therefore, the values which make that:

$R(x)\leq0$

Are in the interval:

$(-2,8\rbrack$