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C3 - Functions

Need some help with 9iii?
Do I put g(x) into f^-1(x) and see where it is not defined?
C3 functions.jpg32.6KB
Original post by Super199
Need some help with 9iii?
Do I put g(x) into f^-1(x) and see where it is not defined?


Yes
Original post by Super199
Need some help with 9iii?
Do I put g(x) into f^-1(x) and see where it is not defined?


Yes.
Reply 3
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Super199
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Original post by TenOfThem
Yes



Original post by morgan8002
Yes.

F^-1(x)= sin^-1(x/2)
So is it sin^-1[(4-2x^2)/2]
Where is it not defined? Not sure how you work that bit out
Original post by Super199
F^-1(x)= sin^-1(x/2)
So is it sin^-1[(4-2x^2)/2]
Where is it not defined? Not sure how you work that bit out


sin⁡−1x\sin^{-1}x is undefined for ∣x∣>1|x|>1
Original post by Super199
F^-1(x)= sin^-1(x/2)
So is it sin^-1[(4-2x^2)/2]
Where is it not defined? Not sure how you work that bit out


You can only take the arc sine of a value between -1 and 1
Reply 6
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Original post by morgan8002
sin⁡−1x\sin^{-1}x is undefined for ∣x∣>1|x|>1


Original post by TenOfThem
You can only take the arc sine of a value between -1 and 1

So do I say it is undefined when (4-2x^2)/2 > 1 and (4-2x^2)/2 <-1?
Original post by Super199
So do I say it is undefined when (4-2x^2)/2 > 1 and (4-2x^2)/2 <-1?

You're on the right lines. You should work out the actual values of x that it is undefined for (solve each quadratic inequality).
Reply 8
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Original post by morgan8002
You're on the right lines. You should work out the actual values of x that it is undefined for (solve each quadratic inequality).

Yeah got it. Cheers to you both :smile:
Reply 9
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Original post by morgan8002
You're on the right lines. You should work out the actual values of x that it is undefined for (solve each quadratic inequality).


Original post by TenOfThem
You can only take the arc sine of a value between -1 and 1

Do you mind helping me with another question :smile:
Part 9ii.
http://www.ocr.org.uk/Images/144592-question-paper-unit-4723-01-core-mathematics-3.pdf

I have to use part 1 but how?
I'm going to have a guess and say 6sin^2(1/2 theta)=2?
Original post by Super199
Do you mind helping me with another question :smile:
Part 9ii.
http://www.ocr.org.uk/Images/144592-question-paper-unit-4723-01-core-mathematics-3.pdf

I have to use part 1 but how?
I'm going to have a guess and say 6sin^2(1/2 theta)=2?

Yes. Let α=12θ\alpha = \frac{1}{2}\theta. Then you can use the identity directly to obtain your guess.
Reply 11
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Original post by morgan8002
Yes. Let α=12θ\alpha = \frac{1}{2}\theta. Then you can use the identity directly to obtain your guess.

Right done, how do you do the last part?
6sin^2(1/3theta)=k
How do I work out k if that is the right equation?
Original post by Super199
Right done, how do you do the last part?
6sin^2(1/3theta)=k
How do I work out k if that is the right equation?

k doesn't have a definite value. It has a range of possible values. The question asks for the set of possible values.

Therefore, you use the maximum and minimum values of 6sin⁡2(13θ)6\sin^2 (\frac{1}{3}\theta) in the given domain as the endpoints of the interval that k can be in.
Reply 13
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Original post by morgan8002
k doesn't have a definite value. It has a range of possible values. The question asks for the set of possible values.

Therefore, you use the maximum and minimum values of 6sin⁡2(13θ)6\sin^2 (\frac{1}{3}\theta) in the given domain as the endpoints of the interval that k can be in.

Sorry but wh
Original post by morgan8002
k doesn't have a definite value. It has a range of possible values. The question asks for the set of possible values.

Therefore, you use the maximum and minimum values of 6sin⁡2(13θ)6\sin^2 (\frac{1}{3}\theta) in the given domain as the endpoints of the interval that k can be in.

mm I don't understand sorry :frown:
Original post by Super199
Sorry but wh
mm I don't understand sorry :frown:


k doesn't have one definite value. It can take a range of values, as the questions says. To find the range of values it can take the minimum is the minimum of 6sin⁡2(13θ)6\sin^2(\frac{1}{3}\theta) in the given domain. The maximum value of k is the maximum of 6sin⁡2(13θ)6\sin^2(\frac{1}{3}\theta) in the given domain.
Reply 15
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Original post by morgan8002
k doesn't have one definite value. It can take a range of values, as the questions says. To find the range of values it can take the minimum is the minimum of 6sin⁡2(13θ)6\sin^2(\frac{1}{3}\theta) in the given domain. The maximum value of k is the maximum of 6sin⁡2(13θ)6\sin^2(\frac{1}{3}\theta) in the given domain.

Domain is -90< theta < 90?
Is the domain now -30< theta/3< 30?
I'm a bit lost
Original post by Super199
Domain is -90< theta < 90?
Is the domain now -30< theta/3< 30?
I'm a bit lost


Yes, that's right. So you need to find the values of 13θ\frac{1}{3}\theta in −30<13θ<30-30<\frac{1}{3}\theta<30 that give the maximum and minimum values of 6sin⁡2(13θ)6\sin^2(\frac{1}{3}\theta), and find these maximum and minimum values.
Reply 17
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Original post by morgan8002
Yes, that's right. So you need to find the values of 13θ\frac{1}{3}\theta in −30<13θ<30-30<\frac{1}{3}\theta<30 that give the maximum and minimum values of 6sin⁡2(13θ)6\sin^2(\frac{1}{3}\theta), and find these maximum and minimum values.

Grr I have been an idiot. Done it! Thanks for the help
Original post by Super199
Grr I have been an idiot. Done it! Thanks for the help

No problem.

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