Question About Test Points

**Past45** · April 3rd 2012, 01:27 PM

This is for Nonlinear inequalities in one variable, via solving the quadratic inequalities. I also have to put the solution set in interval notation.

For example.

X-5/X-7>(or equal to) 0.

First I have to find the denominator that makes the value to 0. So this will be X=7. But also the numerator gets changed to x=5. What my question is, is the test point really needed? Because can't you just look at the solution and see if the number is greater than(or equal to)0?. 5>0 or 7>0. Neither of these are true so the notation set would look like (5,7), right? But also, about the test points, the numbers that get plugged into the variables, are they just random numbers I can pick and choose from?

**Plato** · April 3rd 2012, 01:34 PM

Originally Posted by Past45

This is for Nonlinear inequalities in one variable, via solving the quadratic inequalities. I also have to put the solution set in interval notation.

For example. X-5/X-7>(or equal to) 0.

Please use grouping symbols.
Is it $x-\frac{5}{x-7}\ge 0~?$ OR is it $\frac{x-5}{x-7}\ge 0~?$

**Past45** · April 3rd 2012, 01:37 PM

it's the second one.

**Plato** · April 3rd 2012, 01:42 PM

Originally Posted by Past45

it's the second one.

On what intervals do both numerator and denominator have same sign?

**Past45** · April 3rd 2012, 01:49 PM

Not too sure, I am looking through my book and I do not have a similar problem like that(nor do my notes show a problem with the numerator and denominator with the same symbols), I am going to assume I wrote it wrong the denominator is suppose to be x+5.

So that would make the interval notation be (-∞,-7]U[5,∞)

Originally Posted by Plato

On what intervals do both numerator and denominator have same sign?

**skeeter** · April 3rd 2012, 02:01 PM

$\frac{x-5}{x-7} \ge 0$

critical values are x = 5 and x = 7

test a single value in the each of the following intervals ...

$(-\infty,5]$

$[5,7)$

$(7,\infty)$

since the function is continuous in each interval, if one value in the interval makes the inequality true, then all values in that interval will likewise.

**Plato** · April 3rd 2012, 02:04 PM

Originally Posted by Plato

On what intervals do both numerator and denominator have same sign?

The answer is $(-\infty,5]\cup(7,\infty)$ . WHY?
Why is 5 included and 7 not included?

**Past45** · April 3rd 2012, 02:13 PM

Because the 5 goes through the 7 when the answer has the ∞. I thought the answer would been [5,∞) tbh, since both 5 and 7 are greater than 0. I did not think they would have a union.

But is a test really needed to get the answer?

Originally Posted by Plato

The answer is $(-\infty,5]\cup(7,\infty)$ . WHY?
Why is 5 included and 7 not included?

**Plato** · April 3rd 2012, 02:16 PM

Originally Posted by Plato

The answer is $(-\infty,5]\cup(7,\infty)$ . WHY?
Why is 5 included and 7 not included?

See reply #5.

**Past45** · April 3rd 2012, 02:35 PM

That is true, for this problem it is. I did 3 test points, -3, which the answer came out to be 2 over 5 is greater or equal to 0. Then I did 0, and 5 over 7 is greater than or equal to 0. Then I did 1 and the answer was 2 over 3 is greater or equal to 0.

I have to apologize(seems like I am repeating the question), only thing that is why there is a union when 5 is greater than 0, and we start with (-∞ to 5) U... I would of thought the answer would of been [5,∞).

Originally Posted by skeeter

$\frac{x-5}{x-7} \ge 0$

critical values are x = 5 and x = 7

test a single value in the each of the following intervals ...

$(-\infty,5]$

$[5,7)$

$(7,\infty)$

since the function is continuous in each interval, if one value in the interval makes the inequality true, then all values in that interval will likewise.

**skeeter** · April 3rd 2012, 03:22 PM

Originally Posted by Past45

That is true, for this problem it is. I did 3 test points, -3, which the answer came out to be 2 over 5 is greater or equal to 0. Then I did 0, and 5 over 7 is greater than or equal to 0. Then I did 1 and the answer was 2 over 3 is greater or equal to 0.

I have to apologize(seems like I am repeating the question), only thing that is why there is a union when 5 is greater than 0, and we start with (-∞ to 5) U... I would of thought the answer would of been [5,∞).

-3, 0 , and 1 are in the same interval, $(-\infty, 5)$ ... how about choosing one value in each of the three distinct intervals?

$(-\infty, 5)$ ... $x = -3$

$(5,7)$ ... $x = 6$

$(7, \infty)$ ... $x = 8$

you know the inequality is true at x = 5, because it makes the inequality equal 0. A vertical asymptote exists at x = 7 ... the value of the expression is negative on the left side and positive on the right side of x = 7.

Taking a look at a graph may also help ...

**Past45** · April 3rd 2012, 09:28 PM

That was my question, how do I know which numbers to pick and choose to do a test point? For example if I pick 9, the answer will be 4 over 2 which equals 2>(or equal to) 0 which is true. It's like whatever number I pick is going to be greater than or equal to 0. If it's fraction or a whole number.

Just one last thing, I think this one is somewhat throwing me off because both the num/dem have the same symbol. and the outcome is going to be true for all. So, I am not sure if I will come across another one like this.

BTW, thanks for the helping hand.

Originally Posted by skeeter

-3, 0 , and 1 are in the same interval, $(-\infty, 5)$ ... how about choosing one value in each of the three distinct intervals?

$(-\infty, 5)$ ... $x = -3$

$(5,7)$ ... $x = 6$

$(7, \infty)$ ... $x = 8$

you know the inequality is true at x = 5, because it makes the inequality equal 0. A vertical asymptote exists at x = 7 ... the value of the expression is negative on the left side and positive on the right side of x = 7.

Taking a look at a graph may also help ...

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