The Student Room Group

Finding X without Y

Had a mock today and was asked to show that X=ln2 for the equation Y=(ex-2)2-1
And then asked to find the Y value.
How would this be solved?
Reply 1
Original post by Ky15
Had a mock today and was asked to show that X=ln2 for the equation Y=(ex-2)2-1
And then asked to find the Y value.
How would this be solved?


Huh? Do you mean find Y(ln2)Y(\ln 2)? If so, just plug it in and note that eln2=2e^{\ln 2} = 2
Reply 2
Original post by Zacken
Huh? Do you mean find Y(ln2)Y(\ln 2)? If so, just plug it in and note that eln2=2e^{\ln 2} = 2


No, I meant y=(ex-2)2-1 and I had to show that X=ln2
Reply 3
Original post by Ky15
No, I meant y=(ex-2)2-1 and I had to show that X=ln2


Huh? What does that even mean, are you meant to have yy be something, or do you want to solve for x in terms of y, or what...?
Could you give the full question because that's unsolvable from what you've presented.
Original post by Ky15
Had a mock today and was asked to show that X=ln2 for the equation Y=(ex-2)2-1
And then asked to find the Y value.
How would this be solved?

Just say the answer is in the reals and hope for the best.
Original post by Ky15
Had a mock today and was asked to show that X=ln2 for the equation Y=(ex-2)2-1
And then asked to find the Y value.
How would this be solved?


Some sort of incomplete question if that's all you're given. x=ln(2)x=\ln(2) is the x coordinate of the stationary point though so maybe something to do with that then you can plug it through the original to find the y coordinate and get a set of coordinates for the stationary point.
(edited 7 years ago)
Original post by Ky15
Had a mock today and was asked to show that X=ln2 for the equation Y=(ex-2)2-1
And then asked to find the Y value.
How would this be solved?


There's some information missing from your question. Maybe you were asked to find the x value that minimises y?
Reply 8
That was the full question
Reply 9
Original post by Ky15
That was the full question


It's not a question, it's senseless and either something went wrong on your mock or you simply don't remember the question.
Reply 10
Original post by Ky15
That was the full question

If you're sure then your teacher set you a question that doesn't make sense.
Reply 11
Original post by Mystery.
I think it's asking you to find where function crossed the x-axis which would be when y=0.


The function is clearly everywhere positive (and 1\geq 1) so your reply doesn't make any sense.
(edited 7 years ago)
Reply 12
Original post by Ky15
Had a mock today and was asked to show that X=ln2 for the equation Y=(ex-2)2-1
And then asked to find the Y value.
How would this be solved?


If you're taking Calc I, maybe your teacher asked you to find the critical value of YY? Taking the derivative and setting it equal to 0 would give you X=ln2X=\ln{2}, studying the sign of YY' on either side, and then you'd evaluate Y(ln2)Y(\ln{2}) to find the extrema.
Reply 13
Original post by Zacken
The function is clearly everywhere positive (and 1\geq 1) so your reply doesn't make any sense.


x,ex>0\forall x,e^{ x }>0, substracting two gives an x-intercept, so Y=(ex2)21Y=(e^x-2)^2-1 has an x-intercept (two in fact).
Reply 14
Original post by MartyO
x,ex>0\forall x,e^{ x }>0, substracting two gives an x-intercept, so Y=(ex2)21Y=(e^x-2)^2-1 has an x-intercept (two in fact).


Bleurgh, I read +1+1
Reply 15
Original post by MartyO
If you're taking Calc I, maybe your teacher asked you to find the critical value of YY? Taking the derivative and setting it equal to 0 would give you X=ln2X=\ln{2}, studying the sign of YY' on either side, and then you'd evaluate Y(ln2)Y(\ln{2}) to find the extrema.


You wouldn't need to do all this, of course. You've got a square plus some constant, so the function is minimized when the square is 0.
Reply 16
Original post by Mystery.
I think it's asking you to find where function crossed the x-axis which would be when y=0.


It's not the next part then asked you to find Y when x is ln2 which came out as -1
Reply 17
I found out that I was meant to find dy/dx to see ln2 is a local minimum. Then because it's in the form of completing the square you know Y=-1
Reply 18
Original post by Ky15
I found out that I was meant to find dy/dx to see ln2 is a local minimum. Then because it's in the form of completing the square you know Y=-1

It was clear that there was more to the question than what you posted.

Even though you seemed pretty certain that you had posted the full question :smile:

Quick Reply

Latest