C3 help :)

Watch
Announcements
#1
Need some help with question 6iii.

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

Put them equal to each other and got

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
0
5 years ago
#2
(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

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

0
5 years ago
#3
(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

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:
lies on the line
0
#4
(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
0
#5
(Original post by joostan)
It may be easiest to recall that the solution to:
lies on the line
mm I am confused. Surely I get the same thing but just y instead of x?
0
5 years ago
#6
You should post the problem, i do not know which one from those pages is.
0
5 years ago
#7
(Original post by Super199)
mm I am confused. Surely I get the same thing but just y instead of x?
No, since at we have both and you can pick either of the latter two equations and solve that, which is simpler.
0
5 years ago
#8
(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
0
#9
(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 we have both and 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
0
5 years ago
#10
(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?
0
#11
(Original post by joostan)
No worries
At P what can you say about the second derivative?
less than 0
0
5 years ago
#12
(Original post by Super199)
less than 0
Nearly, the gradient takes the maximum value, not the function, so ?
0
#13
(Original post by joostan)
Nearly, the gradient takes the maximum value, not the function, so ?
=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?
0
5 years ago
#14
(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:
.
to find the gradient of the tangent.
1
#15
(Original post by joostan)
Yes you'll need to sub e in to:
.
to find the gradient of the tangent.
Yh got it! Thanks again
1
#16
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
0
X

new posts
Back
to top
Latest
My Feed

Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

See more of what you like onThe Student Room

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

Poll

Join the discussion

Yes (194)
60.06%
No (71)
21.98%
Not sure (58)
17.96%