solve for range of k. | Mathematics | Page 1

karmic glen Feb 25, 2023, 2:33 PM

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Someone help I am struggling a lot thank you

celest cloudBOT Feb 25, 2023, 2:33 PM

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glad void Feb 25, 2023, 2:42 PM

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"y can assume all real values", means that the range of y should be R (the reals). Do you know how to determine the range of that function?
(Correction I said "domain" once, I meant range)

karmic glen Feb 25, 2023, 3:40 PM

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@glad void use the discriminant?

glad void Feb 25, 2023, 3:42 PM

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What have you tried? Or where are you currently?

karmic glen Feb 25, 2023, 3:42 PM

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i first factorised the denominator

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and found that x cannot be 1 or -3, or else the denominator is zero and y would not be a real number

glad void Feb 25, 2023, 3:43 PM

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Ok good, so you know the domain of y

karmic glen Feb 25, 2023, 3:43 PM

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yes

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then i found the discriminant of the numerator

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but idk if thats the way to go

glad void Feb 25, 2023, 3:45 PM

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That will give you roots (zeroes) of the function, but I don't think it will help with the problem

lethal granite Feb 25, 2023, 3:45 PM

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sorry for interrupting, but i think if we make discriminant greater than or equal to zero then that should do it

karmic glen Feb 25, 2023, 3:46 PM

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lethal granite sorry for interrupting, but i think if we make discriminant greater than or equa...

well i did that and got k<= 0 or k>=4, but idt its complete yet

glad void Feb 25, 2023, 3:46 PM

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You want to determine the range of y in (-inf, -3), (-3, 1) and (1, inf)

karmic glen Feb 25, 2023, 3:47 PM

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glad void You want to determine the range of y in (-inf, -3), (-3, 1) and (1, inf)

oh

glad void Feb 25, 2023, 3:48 PM

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Those ranges together will give you the range of y

karmic glen Feb 25, 2023, 3:48 PM

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glad void You want to determine the range of y in (-inf, -3), (-3, 1) and (1, inf)

why is there infinity here tho, im confused a little

lethal granite Feb 25, 2023, 3:49 PM

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karmic glen why is there infinity here tho, im confused a little

y can be anything except -3 and 1 so that's why

glad void Feb 25, 2023, 3:49 PM

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R = (-inf, inf), removing the poles X = -3 and x = 1, you end up with that domain

karmic glen Feb 25, 2023, 3:50 PM

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ok

glad void Feb 25, 2023, 3:51 PM

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And for each of the subdomains you should check what the function does at the boundary, and what it does in between

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So take the limit from both directions for each of the boundaries. Thinking about this, having determined the zeroes is helpful here. Maybe a pole can be removed

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For the insides you want to check for minima and maxima, unless you already know the range is (-inf, inf)

karmic glen Feb 25, 2023, 3:54 PM

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could you explain what a pole and a subdomain is? I am not sure

glad void Feb 25, 2023, 3:56 PM

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Poles are the zeroes of the denominator, when it is equal to zero. So the singularities of the function. x = -3 and x = 1

karmic glen Feb 25, 2023, 3:57 PM

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got it

karmic glen Feb 25, 2023, 3:58 PM

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glad void You want to determine the range of y in (-inf, -3), (-3, 1) and (1, inf)

are these coordinates?

glad void Feb 25, 2023, 3:58 PM

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No it's a set

karmic glen Feb 25, 2023, 3:58 PM

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oh

glad void Feb 25, 2023, 3:59 PM

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(-inf, -3) is the set of all numbers between -inf and -3

karmic glen Feb 25, 2023, 3:59 PM

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icic

glad void Feb 25, 2023, 4:00 PM

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karmic glen could you explain what a pole and a subdomain is? I am not sure

Not sure if subdomain is the correct wording but I meant the 3 sets that form the domain

karmic glen Feb 25, 2023, 4:00 PM

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ok

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so i need to solve 3 instances for y

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(-inf, -3), (-3, 1) and (1, inf)

glad void Feb 25, 2023, 4:02 PM

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You don't really need to solve, calculate what the function does at those boundaries

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Well you do need to solve for which k that is

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Is this a bonus exercise?

karmic glen Feb 25, 2023, 4:03 PM

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yes

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its more challenging than i thought

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cuz we just got into the topic of the nature of roots and using the discriminant

glad void Feb 25, 2023, 4:05 PM

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You can use the information from the discrimant

karmic glen Feb 25, 2023, 4:06 PM

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i got k<= 0 or k>=4 using the discriminant

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for the numerator to have a real root

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but here the root does not need to be real, just the value of y

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in this case can't k be anything, say (-inf to inf) so long as x isnt equal to -3 or 1, y will have a real value

glad void Feb 25, 2023, 4:13 PM

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Well what you do know is, if there are no roots, y is never zero and thus the range is not R

karmic glen Feb 25, 2023, 4:15 PM

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oh yea if there are no roots k is not R

glad void Feb 25, 2023, 4:15 PM

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Yes

karmic glen Feb 25, 2023, 4:16 PM

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i got k ∈ (-∞, -1] ∪ [9/2, ∞)

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i referred to a worksheet involving solution set

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nvm scratch that

glad void Feb 25, 2023, 4:18 PM

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Oh

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Well you do know it isn't (0, 4)

karmic glen Feb 25, 2023, 4:19 PM

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yea

glad void Feb 25, 2023, 4:20 PM

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Well now first check where the poles go to

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Sorry, I'm not really sure what is the best way to go forward

karmic glen Feb 25, 2023, 4:23 PM

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the poles are the values when you set the discriminant to zero right

glad void Feb 25, 2023, 4:24 PM

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No when the denominator goes to zero and thus y goes to -inf or inf, unless it is a removable singularity

karmic glen Feb 25, 2023, 4:24 PM

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oh

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wait but isnt the solution just k<= 0 or k>=4 since the numerator needs to have real roots

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and the discriminant of the numerator is k(k-4) >= 0

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for the roots to be real

glad void Feb 25, 2023, 4:27 PM

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Hm

karmic glen Feb 25, 2023, 4:27 PM

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and ultimately for y to be real

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ok i think i need to sleep now its getting quite late

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ill attempt this question tmr

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thx for the help @glad void

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and @lethal granite

glad void Feb 25, 2023, 4:29 PM

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If you have root, and the singularities goes to -inf and inf, then R has the whole range

glad void Feb 25, 2023, 4:30 PM

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karmic glen thx for the help <@968981177002246216>

I will try to work it out

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Have a good night

karmic glen Feb 25, 2023, 4:30 PM

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u too

glad void Feb 25, 2023, 5:40 PM

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These are all asymptotes

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Maybe I made some sign mistakes...

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Yeah the k < and > 9/2 should be flipped

glad void Feb 25, 2023, 6:16 PM

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Also in the graph for x < -3 or x > 1 it is still possible for it to go through 1, but eventually it will go to the asymptote y = 1 again.

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glad void Feb 25, 2023, 6:56 PM

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To summarize:

function is a rational polynomial
degree of numerator and denominator are the same and we have a horizontal asymptote at y = 1/1 = 1
function has to intersect y = 1 or the range cannot be R
since we have a horizontal asymptotes at +-infinity we have to look at the poles, the vertical asymptotes. Determine these, they are x = -3 and x = 1.
If the limits between them (positive left, negative right) have opposite signs, the range will be (-inf, inf). Determine these ks. (1)
For the other cases the intersection with y = 1 has to be in the regions that should span the whole range. Determine for which k and x, f(x) intersects with y = 1. (a)
One of these cases is that the negative left and positive right asymptotes have opposite limits. Normally you would need to check the ks and determine if the intersection (a) is at x < -3 or x > 1. (*Wait this is not true, the left and right always span everything except y = 1, so it's enough for there to be a intersection) But it turns out these k are the same as (1).
Another is when the negative left and positive right limits have the same sign but opposite to the positive left and negative right limits. For this the intersection (a) has to lay between (-3, 1). It turns out that this isn't the case, the only intersections take place in the left and right region when - 1/2 <= k <= 9/2

So your answer will be the ks determined at (1)

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I guess that is it, can someone validate? 👼

glad void Feb 25, 2023, 9:48 PM

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(1) the left limits flip at k = -1/2, check which is which. The same for the right limits with k = 9/2.
Turns out the middle region spans R when k < - 1/2 or k > 9/2. (THIS IS THE ANSWER)

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(a) when determining the intersection, it turns out there are no intersections for k in {-1/2, 2, 9/2} the first and last come from x in {1, -3}. The intersection is at x = (-k - 3)/(k -2).

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Well that was fun, learned some algebra, about rational polynomials and it's asymptotes. Not sure how you should do this when you just learn about the zeroes. Unless there is a big shortcut...

#solve for range of k.