Reasoning behind this C4 maths question?

Hey, I've put a picture of the question (you might have to click on it to enlarge).
Screen Shot 2016-02-14 at 12.31.35.png

Anyway, I can do part a). I can do like half of part b). I set dy/dx = 0, then found x=2y^2 and inserted this into the equation of the curve, and got: 9y^4 + 9 = 0. Now when i was doing the paper, i actually went on and said 'y^4 = -1, not possible therefore no point on C' but this is not correct. On the markscheme they also get 9y^4 +9 = 0 BUT then they say, '9y^4 + 9 > 0 for any real y and thus no such point exists'. What does this mean? Can someone explain it in simpler terms? I know it's probably something simple but I don't understand the 'real y' thing.
ALSO, i don't really get the reasoning behind the whole point of doing part b) either. i kind of just guessed what to do if i'm honest but i don't really no why (sorry if this last bit doesn't make sense haha. the main thing is what i've written above).
Thanks in advance! :biggrin:

(edited 8 years ago)

Reply 1

Zacken

22

Original post by Phoebus Apollo

Hey, I've put a picture of the question (you might have to click on it to enlarge).
Screen Shot 2016-02-14 at 12.31.35.png

Anyway, I can do part a). I can do like half of part b). I set dy/dx = 0, then found x=2y^2 and inserted this into the equation of the curve, and got: 9y^4 + 9 = 0. Now when i was doing the paper, i actually went on and said 'y^4 = -1, not possible therefore no point on C' but this is not correct. On the markscheme they also get 9y^4 +9 = 0 BUT then they say, '9y^4 + 9 > 0 for any real y and thus no such point exists'. What does this mean? Can someone explain it in simpler terms? I know it's probably something simple but I don't understand the 'real y' thing.
Thanks in advance! :biggrin:

What you've done/said is entirely correct and equivalent to what the markscheme has done.

But essentially,

9y^4 + 9

is always greater than zero, so if you imagine it's graph, it never becomes 0 or gets lower than zero, essentially, it always lies above the x-axis, and hence can never =0 because =0 means that it intersects the x-axis. :smile:

Does that help or do you need more of an explanation? Feel free to ask more! :biggrin:

Reply 2

Zacken

22

Original post by Phoebus Apollo

ALSO, i don't really get the reasoning behind the whole point of doing part b) either. i kind of just guessed what to do if i'm honest but i don't really no why (sorry if this last bit doesn't make sense haha. the main thing is what i've written above).
Thanks in advance! :biggrin:

Just seen your edit: I'm not entire sure what you mean by not getting the reasoning of part(b). You want to prove that no such point exists. This is not apparent, so you do some valid mathematical working till you get to a stage where it is apparent that no such point exists.

Reply 3

PLM98

6

You just have to explain that it is impossible to get 0 because whenever you do something to the power of 4 it is always going to be positive, and then you times it by 9 and add 9 and its still positive. There is no y value that would give you an answer of 0 or a negative answer. You just have to state that 9y^4 + 9 > 0 for any real y value as the mark scheme says, and therefore there is no point at which dy/dx=0. Real y means a real number (ie any number between minus infinity and plus infinity), so there is no real number that when substituted for y would give you an answer of 0, hence no such point exists.

Reply 4

Phoebus Apollo

OP

6

Original post by Zacken

Just seen your edit: I'm not entire sure what you mean by not getting the reasoning of part(b). You want to prove that no such point exists. This is not apparent, so you do some valid mathematical working till you get to a stage where it is apparent that no such point exists.

Thanks for your reply! I kind of get what you mean, but how does this prove that there is no point at which the gradient (dy/dx) = 0?
:smile:

Reply 5

Zacken

22

Original post by Phoebus Apollo

Thanks for your reply! I kind of get what you mean, but how does this prove that there is no point at which the gradient (dy/dx) = 0?
:smile:

Because your working has shown that:

dy/dx = 0 if and only if x = 2y^2

and

x=2y^2 if and only if 9y^4 + 9 = 0

so, this can be concatenated into:

dy/dx = 0 if and only if 9y^4 + 9 =0

So since 9y^4 + 9 is NOT zero, then dy/dx is also not 0.

(edited 8 years ago)

Reply 6

the bear

20

dy/dx = 0 <===> the top of the fraction = 0

your version is valid.

Reply 7

Phoebus Apollo

OP

6

Original post by Zacken

Because your working has shown that:

dy/dx = 0 if and only if x = 2y^2

and

x=2y^2 if and only if 9y^4 + 9 = 0

so, this can be concatenated into:

dy/dx = 0 if and only if 9y^4 + 9 =0

So since 9y^4 + 9 is NOT zero, then dy/dx is also not 0.