Solving Rational Inequalities Algebraically

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beachpanda
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Hi - I've attached a section from my Further Maths textbook which explains how to solve rational inequalities algebraically.

But I'm struggling to see how we know whether to use a less than/greater than sign or a less than or equal to sign etc?

It's only from looking at the graph of the function that I can see why it's x ≤ - 2 rather than x < - 2 for example.

Is anyone able to help?
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mqb2766
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(Original post by beachpanda)
Hi - I've attached a section from my Further Maths textbook which explains how to solve rational inequalities algebraically.

But I'm struggling to see how we know whether to use a less than/greater than sign or a less than or equal to sign etc?

It's only from looking at the graph of the function that I can see why it's x ≤ 2 rather than x < 2 for example.

Is anyone able to help?
You realize why they're interested in those 4 intervals?
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ghostwalker
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(Original post by beachpanda)
Hi - I've attached a section from my Further Maths textbook which explains how to solve rational inequalities algebraically.

But I'm struggling to see how we know whether to use a less than/greater than sign or a less than or equal to sign etc?

It's only from looking at the graph of the function that I can see why it's x ≤ 2 rather than x < 2 for example.

Is anyone able to help?
Can x=2? Look at the original function.
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beachpanda
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(Original post by ghostwalker)
Can x=2? Look at the original function.
Sorry typo, I meant:

(Original post by beachpanda)
It's only from looking at the graph of the function that I can see why it's x ≤ - 2 rather than x < - 2 for example.
(Original post by mqb2766)
You realize why they're interested in those 4 intervals?
I think so yeah as these are the points where there's roots, local maxima and asymptotes?
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mqb2766
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(Original post by beachpanda)
Sorry typo, I meant:




I think so yeah as these are the points where there's roots, local maxima and asymptotes?
Those are the fancy names, but for this problem its where one of the factors flips signs.
If you multiplying or dividing, you need an odd number of factors to be negative for the result < 0
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ghostwalker
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(Original post by beachpanda)
It's only from looking at the graph of the function that I can see why it's x ≤ - 2 rather than x < - 2 for example.
When dealing with the critical points, we can go back to the original inequality and check. So, sub in x=-2, and does it satisfy the inequality?

You would need to do that as the original inequality, may be strict, "<", or not ,"<=". If it's strict, then x=-2 doesn't satisfy the original inequality, but here it's not strict, and it does satisfy it, and hence we include it in the solution set.
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beachpanda
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(Original post by ghostwalker)
When dealing with the critical points, we can go back to the original inequality and check. So, sub in x=-2, and does it satisfy the inequality?

You would need to do that as the original inequality, may be strict, "<", or not ,"<=". If it's strict, then x=-2 doesn't satisfy the original inequality, but here it's not strict, and it does satisfy it, and hence we include it in the solution set.
That makes a lot more sense, thanks!
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