C3 Differentiation / Set of Values help

Watch
Dggj_19
Badges: 11
Rep:
?
#1
Report Thread starter 2 years ago
#1
Name:  7q.JPG
Views: 45
Size:  31.8 KB

The answer for (a) is as follows: g(x) = (22x^2 - 2)
For (b), I understand that to find a set of values for which dy/dx ≥ 0, you have to equate the two factors of dy/dx, those being
(22x^2 - 2) and (x^2 - 1)^4 respectively, as equalling 0. this gives +/- 1 and +/- √(1/11). Why is +/- √(1/11) the correct set of values and not +/- 1? it's not a range, but two distinct values at which the gradient is 0, so it's not as if one set of values can be included in the other. Just confused as to why it's specifically one and not the other. thanks for any help!
Last edited by Dggj_19; 2 years ago
0
reply
RDKGames
Badges: 20
Rep:
?
#2
Report 2 years ago
#2
(Original post by Dggj_19)

The answer for (a) is as follows: g(x) = (22x^2 - 2)
For (b), I understand that to find a set of values for which dy/dx ≥ 0, you have to equate the two factors of dy/dx, those being
(22x^2 - 2) and (x^2 - 1)^4 respectively, as equalling 0. this gives +/- 1 and +/- √(1/11). Why is +/- √(1/11) the correct set of values and not +/- 1? it's not a range, but two distinct values at which the gradient is 0, so it's not as if one set of values can be included in the other. Just confused as to why it's specifically one and not the other. thanks for any help!
We have (22x^2-2)(x^2-1)^4 \geq 0 when

(*) 22x^2 - 2 \geq 0
AND
(**) (x^2 -1)^4 \geq 0.

The first inequality (*) is true when x \leq -\dfrac{1}{\sqrt{11}} or x \geq \dfrac{1}{\sqrt{11}}.

The second inequality (**) is true for all x; it is wrong to say it's only true for  x \leq -1 or x \geq 1. Take 0 for instance, clearly (0^2 - 1)^4 = 1 > 0 so the second inequality is true.

Since the (**) does not constrain our values of x, but the first inequality does, then we ignore the second inequality.


We want values of x for which both inequalities hold, and clearly the second inequality doesn't care what we choose as it always holds, so we only need to ensure that (*) holds. Hence the values from it constitute our answer.
Last edited by RDKGames; 2 years ago
0
reply
Dggj_19
Badges: 11
Rep:
?
#3
Report Thread starter 2 years ago
#3
(Original post by RDKGames)
We have (22x^2-2)(x^2-1)^4 \geq 0 when

(*) 22x^2 - 2 \geq 0
AND
(**) x^2 -1 \geq 0.

The first inequality (*) is true when x \leq -\dfrac{1}{\sqrt{11}} or x \geq \dfrac{1}{\sqrt{11}}.

The second inequality (**) is true for all x; it is wrong to say it's only true for  x \leq -1 or x \geq 1. Take 0 for instance, clearly (0^2 - 1)^4 = 1 > 0 so the second inequality is true.

Since the (**) does not constrain our values of x, but the first inequality does, then we ignore the second inequality.


We want values of x for which both inequalities hold, and clearly the second inequality doesn't care what we choose as it always holds, so we only need to ensure that (*) holds. Hence the values from it constitute our answer.
This makes sense, and I think I understand now. Thank you!
0
reply
RDKGames
Badges: 20
Rep:
?
#4
Report 2 years ago
#4
(Original post by Dggj_19)
This makes sense, and I think I understand now. Thank you!
Small error on (**) but I fixed it now.

Hopefully it's clear to you why raising something to the 4th power will always make it \geq 0 no matter the input.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

How are you feeling about starting university this autumn?

Really excited (74)
22.7%
Excited but a bit nervous (148)
45.4%
Not bothered either way (38)
11.66%
I'm really nervous (66)
20.25%

Watched Threads

View All