Super199
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Need some help with question 6iii.

I got the inverse function to be (x-1)^2

Put them equal to each other and got  x^2-2x-\sqrt x=0

But then I had troubles solving that if I am doing it right?

I tried letting y = square root of x but that just gave a quartic which I dunno how to solve.

Any help would be appreciated.

http://www.ocr.org.uk/Images/74918-s...rk-schemes.pdf
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TenOfThem
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(Original post by Super199)
Need some help with question 6iii.

I got the inverse function to be (x-1)^2

Put them equal to each other and got  x^2-2x-\sqrtx=0


If you put y = root x you get a quartic not a quadratic

y^4 - 2x^2 - y = 0
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joostan
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(Original post by Super199)
Need some help with question 6iii.

I got the inverse function to be (x-1)^2

Put them equal to each other and got  x^2-2x-\sqrt x=0

But then I had troubles solving that if I am doing it right?

I tried letting y = square root of x but that just gave a quartic which I dunno how to solve.

Any help would be appreciated.

http://www.ocr.org.uk/Images/74918-s...rk-schemes.pdf
It may be easiest to recall that the solution to:
f(x)=f^{-1}(x) lies on the line y=x
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Super199
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(Original post by TenOfThem)
This is correct but, if you put y = root x you get a quartic not a quadratic
Yh that is what I wrote
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Super199
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(Original post by joostan)
It may be easiest to recall that the solution to:
f(x)=f^{-1}(x) lies on the line y=x
mm I am confused. Surely I get the same thing but just y instead of x?
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DiaanaE
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You should post the problem, i do not know which one from those pages is.
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joostan
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(Original post by Super199)
mm I am confused. Surely I get the same thing but just y instead of x?
No, since at f(x)=f^{-1}(x) we have both x=f(x) and x=f^{-1}(x) you can pick either of the latter two equations and solve that, which is simpler.
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TenOfThem
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(Original post by Super199)
mm I am confused. Surely I get the same thing but just y instead of x?
As joostan points out

Since you are on the line y=x

x = f(x)

or indeed

x=f^-1(x)

will be easy to solve
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Super199
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(Original post by TenOfThem)
As joostan points out

Since you are on the line y=x

x = f(x)

or indeed

x=f^-1(x)

will be easy to solve
(Original post by joostan)
No, since at f(x)=f^{-1}(x) we have both x=f(x) and x=f^{-1}(x) you can pick either of the latter two equations and solve that, which is simpler.
oo that is a lot easier. Thanks

I also need help with 8ii if you guys don't mind
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joostan
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(Original post by Super199)
oo that is a lot easier. Thanks

I also need help with 8ii if you guys don't mind
No worries
At P what can you say about the second derivative?
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Super199
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(Original post by joostan)
No worries
At P what can you say about the second derivative?
less than 0
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joostan
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(Original post by Super199)
less than 0
Nearly, the gradient takes the maximum value, not the function, so \dfrac{d^2y}{dx^2}=...?
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Super199
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(Original post by joostan)
Nearly, the gradient takes the maximum value, not the function, so \dfrac{d^2y}{dx^2}=...?
=0

So 2-2lnx= 0
2(1-ln x) =0
ln x =1
x= e

Then just sub e into (ln x)^2 for y-coordinate

and use y-y1=m(x-x1)? Is that how you do it?
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joostan
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(Original post by Super199)
=0

So 2-2lnx= 0
2(1-ln x) =0
ln x =1
x= e

Then just sub e into (ln x)^2 for y-coordinate

and use y-y1=m(x-x1)? Is that how you do it?
Yes you'll need to sub e in to:
\dfrac{dy}{dx}.
to find the gradient of the tangent.
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Super199
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(Original post by joostan)
Yes you'll need to sub e in to:
\dfrac{dy}{dx}.
to find the gradient of the tangent.
Yh got it! Thanks again
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Super199
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Does anyone mind helping me with 9ii from the same paper. I got a to be 1/2pi. But dont understand how to find x? May seem a bit daft but any help would be appreciated
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