Polynominal inequality: (x + 7) (x + 1) >= 0

[Gadsilla]

Polynominal inequality. I'm struggling with the pattern of doing it. The explanations make sense, but I don't know the way to go forward with it.
(x+7)(x+1)>=0

 

[pka]

Polynominal inequality. I'm struggling with the pattern of doing it. The explanations make sense, but I don't know the way to go forward with it.
(x+7)(x+1)>=__?__

If you don't fill that blank then you have posted a nonsense question.

 

[Gadsilla]

If you don't fill that blank then you have posted a nonsense question.

Sorry I thought I typed in the 0.

 

[pka]

Sorry I thought I typed in the 0.

For \(\displaystyle (x+7)(x+1)> 0\) means that both of \(\displaystyle (x+7)~\&~(x+1)\) must have the same sign. Do you understand why?
If \(\displaystyle x> -1\) then they both are positive. WHY?
If \(\displaystyle x< -7\) then they both are negative. WHY? Therefore their product is positive.

 

[JeffM]

Polynominal inequality. I'm struggling with the pattern of doing it. The explanations make sense, but I don't know the way to go forward with it.
(x+7)(x+1)>=0

For an inequality involving a continuous function, first solve the related equality.

\(\displaystyle f(x) = (x + 7)(x + 1) = 0 \text { if and only if } x = -\ 7 \text { or } x = -\ 1.\)

Do you see that?

So if x is neither - 7 nor - 1, f(x) is not zero, which means f(x) is either positive or negative. Is that clear?

Now f(x) is continuous, which means that f(a) is close to f(b) if a and b are close enough to each other. That is turn means that the function does not jump from positive to negative without passing through zero. Because we he have two values where f(x) = 0, they divide the real numbers into three contiguous groyps, numbers < - 7, the numbers > - 7 but < - 1, and the numbers greater than - 1. Every number in one of those groups will have the same sign. So pick a number that is easy to work with in a group and see what sign you get when you pkug that number into the function.

\(\displaystyle -\ 8 < -\ 7 \text { and } f(-\ 8) = (-\ 8 + 7)(-\ 8 + 1) = (-\ 1)(-\ 7) = 7 > 0 \implies \\

f(x) > 0 \text { if } x < -\ 7.\)

Now test the other regions. when that is done, put all the information together.

This technique will work for any inequality involving a continuous function.

 

[Gadsilla]

For an inequality involving a continuous function, first solve the related equality.

\(\displaystyle f(x) = (x + 7)(x + 1) = 0 \text { if and only if } x = -\ 7 \text { or } x = -\ 1.\)

Do you see that?

So if x is neither - 7 nor - 1, f(x) is not zero, which means f(x) is either positive or negative. Is that clear?

Now f(x) is continuous, which means that f(a) is close to f(b) if a and b are close enough to each other. That is turn means that the function does not jump from positive to negative without passing through zero. Because we he have two values where f(x) = 0, they divide the real numbers into three contiguous groyps, numbers < - 7, the numbers > - 7 but < - 1, and the numbers greater than - 1. Every number in one of those groups will have the same sign. So pick a number that is easy to work with in a group and see what sign you get when you pkug that number into the function.

\(\displaystyle -\ 8 < -\ 7 \text { and } f(-\ 8) = (-\ 8 + 7)(-\ 8 + 1) = (-\ 1)(-\ 7) = 7 > 0 \implies \\

f(x) > 0 \text { if } x < -\ 7.\)

Now test the other regions. when that is done, put all the information together.

This technique will work for any inequality involving a continuous function.

So,

(-7+7)(-7+1)
(0)(6)=0
(-1+7)(-1+1)
(6)(0)=0

Solution: (-7)(-1)
?

Edit: Nevermind,

(-∞, -7)(-1 ,∞)

is that right ?
My reasoning is that because higher than -1 and it'll go above 0, and below -7 and it'll go above 0 as well into infinity.
I'm so happy it makes sense

 

[pka]

(-∞, -7)(-1 ,∞)
is that right ?
My reasoning is that because higher than -1 and it'll go above 0, and below -7 and it'll go above 0 as well into infinity.
I'm so happy it makes sense

Yes that is correct.
However, the answer should be written as \(\displaystyle (-\infty,-7)\cup(-1,\infty)\).

 

[Gadsilla]

Yes that is correct.
However, the answer should be written as \(\displaystyle (-\infty,-7)\cup(-1,\infty)\).

What is the U for, to show that it isn't +,÷,-,* ?

 

[Dr.Peterson]

What is the U for, to show that it isn't +,÷,-,* ?

The U means "union": the solution set is the union of those two intervals, meaning that x can be in either of them. Writing two intervals next to one another has no meaning.[FONT=MathJax_Main]
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