why rational inequalities are tricky!
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- Опубліковано 10 жов 2021
- Solving a rational inequality could be tricky because w cannot just multiply both sides by the common denominator and treat it like a linear inequality. We must make sure to work with how the rational inequality is written originally. Pay attention to the restriction of rational expressions. For more precalculus tutorials, you can subscribe here 👉 bit.ly/just_calc
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Just calculus.
you can multiply the equation by x+3 and consider two cases where in case 1, x+3 is positive and in case 2, x+3 is negative. In case of negative you can flip the sign of inequality.
Simply because the sign of the quantity you multiply by in an inequality matters. We could instead multiply by (x+3)^2 which is always positive so the inequality becomes quadratic, and has the same solutions as the given rational inequality.
I’m in my first year of a math major and i have literally never seen it done in this way, very cool
Hello ... And ... Thank you
Didn't know you had this channel, so I subscribed. Love these kinds of problems, but I did not know about this kind of difference.
There is a much simpler method for such inequalities. Just treat them as fractions a/b (where b≠0). When is this fraction greater than zero? When product of a and b is greater than zero.
Therefore we can just write (x-2)(x+3)≥0, keeping in mind that x≠-3. Then proceed like in quadratic or polynomial inequalities to get the same result.
It basically that you can’t multiply or divide by a negative without changing the inequality (making the current one untrue), so let’s say we can only multiply by x+3 when it is positive, which means x > -3 and as u did we get x>=2 which means for all x > -3 ONLY x>= 2 work, i.e. everything inbetween does not, we must now check when x-3 is negative which is when x < -3 which it is then clear that the numerator is negative for akk x < -3 and thus -/- = + which satisfies inequality so all x < -3 work too.
I rather assume x + 3 > 0 and work out the inequality. Then repeat for x + 3 < 0
if x + 3 > 0 , ( x + 3 ) * [ ( 2x + 1 ) / ( x + 3 ) ] > or = 1 ( x + 3 )
2x + 1 > or = x + 3
x > or = 2
and x > - 3
if x + 3 < 0, ( x + 3 ) * [ ( 2x + 1 ) / ( x + 3 ) ] < or = 1 * ( x + 3 )
2x + 1 < or = x + 3
x < or = 2
and x < - 3
so x belongs to ( - inf, - 3 ) U [ 2, inf )
But how do you decide which two of the four answers to discard?
My number one tip while solving linear inequality
If it's positive like a whole square then you can cross multiply it
Thank you so much for all the content you provided 🖤❤.
but can I ask you to do a video about absolute function 🙏 , i mean finding limit/solve and some examples
I feel selfish to ask that so if you feel like doing it I will appreciate it so much 🙂
Can you integrate square root of cos2x/cosx
you can multiply both sides by (x+3)^2, then you will be able to solve it like a quadratic inequality
because k^2 >= 0, so it's a legal move
What I've always wondered is if there is a way to solve these without test points. It always seemed so inelegant.
But test point method is already powerful enough and simpler.
@@HakingMC I get that it this method is probably simpler. But that doesn't change my desire to find a method that doesn't require guess and check at the end. Surely there is some underlying logic.
And understanding the underlying logic is what makes math fun.
@@ZipplyZane finding the sign of a division of polynomials is equivalent to finding the sign of their products (division and multiplication follow the same sign rules +/-=- for example).With that out of the way,when we have a product of polynomials( (x+3)(x-2) )the whole expression retains its sign between and outside of roots as a continuous function,you cant have a change in sign without crossing through a root in other words.(that doesn't mean tho that whenever you cross a root you necessarily change sign, that's why we check in the first place)(think of x^2 ,x=0 is a root of the polynomial and when you go from negative x's to positives ones, x^2 is consistently positive,with the exception of 0)
multiply both sides by (x+3)^2
@@GhostHawk272 That gets me to
(x+3)(x-2) >= 0
But how do I then get x >= -3 or x
My gut is telling me no because why would math be that easy?
😆
Thanks I'm writing admaths in a week and this is probably going to save me an extra 5 marks
This could have been done on your fast channel, Quick Attack Style.
In the first one, x cannot equal 3. In the second one, x can equal 3. Therefore, they are not the same. I'm done in like 5 seconds.
I thought you do these inequalities by considering x0. Then flip the sign when doing x
My thoughts before watching the video: so, I think the answer is no since x+3 hypothetically could be negative depending on the value of x, meaning they describe different inequalities, I think
Edit: ok
Who else comments while they’re watching the video as opposed to after watching the video?
Ah yes calculus
Just try putting x = -4
You will see that they are not the same