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Further Maths GCSE AQA Paper 1 (Unofficial mark scheme)

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Original post by NiamhM1801
Oh, also I didn't get +75x for the cubic, I got +25x. I'm probably wrong though haha


Hello there. I require help with a question and you seem like the right person for this.
Find the turning point of the following functions:
1) y = 0.5x^2 - 2x
I first found dy/dx: x - 2 = 0 thus x = 2.
I substituted this into the original equation to find y: 0.5(4) -2(2) = 0 so my
coordinates are (2,0)
I found d2y/dx^2 to see if it is a maximum or minimum point: x = 0 which is a
positive value so I put down it is a minimum point. However, the correct
answer is that there are NO turning points for the above equation. Please
enlighten me ASAP. This is driving me nuts. Cheers!
Original post by Wolfram Alpha
Hello there. I require help with a question and you seem like the right person for this.
Find the turning point of the following functions:
1) y = 0.5x^2 - 2x
I first found dy/dx: x - 2 = 0 thus x = 2.
I substituted this into the original equation to find y: 0.5(4) -2(2) = 0 so my
coordinates are (2,0)
I found d2y/dx^2 to see if it is a maximum or minimum point: x = 0 which is a
positive value so I put down it is a minimum point. However, the correct
answer is that there are NO turning points for the above equation. Please
enlighten me ASAP. This is driving me nuts. Cheers!


Well firstly, there is a stationary point as it is a quadratic graph and they all have one
Through differentiation you have found that there is a stationary point when x = 2, which is correct.
By substitution into the original equation you should have got 4 - 4 = 0 which you did.
Therefore the stationary point is (2,0). That's all you need to do - it didn't ask for classification so you didn't need to go any further than this, however since it is a regular quadratic graph (not upside down) it's pretty obvious it could only be a minimum point anyway.

EDIT: Sorry I'm wrong - you should have got 2 - 4 = -2 for the y value, therefore the coordinates of the stationary point are (2,-2)
(edited 7 years ago)
Original post by NiamhM1801
Well firstly, there is a stationary point as it is a quadratic graph and they all have one
Through differentiation you have found that there is a stationary point when x = 2, which is correct.
By substitution into the original equation you should have got 4 - 4 = 0 which you did.
Therefore the stationary point is (2,0). That's all you need to do - it didn't ask for classification so you didn't need to go any further than this, however since it is a regular quadratic graph (not upside down) it's pretty obvious it could only be a minimum point anyway.


The correct answer is that there is no turning point so how can it be a local minimum?
Original post by Wolfram Alpha
The correct answer is that there is no turning point so how can it be a local minimum?


Where is this question from? This curve definitely has one
Anyone help with this question??
Find the point of inflection of the curve
y=x^5 + x^4
Original post by Ano123
Anyone help with this question??
Find the point of inflection of the curve
y=x^5 + x^4


Find the second derivative, set it to 0 and solve for x.
Original post by Chittesh14
Find the second derivative, set it to 0 and solve for x.


I did exactly the same thing and got (0.6, 0.78) but are you allowed to use the second derivative method because it doesnt say anything about that in my textbook nor in the mark schemes.
Original post by Darshan98
I did exactly the same thing and got (0.6, 0.78) but are you allowed to use the second derivative method because it doesnt say anything about that in my textbook nor in the mark schemes.


How did you get that? This is what I would have done:

dy/dx = 5x^4 + 4x^3
d^2y/dx^2 = 20x^3 + 12x^2 = 0
x^2(20x + 12) = 0
x = 0 OR
x = -12/20 = - 6/10 = - 3/5
Original post by Darshan98
Sorry my bad. yeah it is -0.6 and -0.08. I keep making so many silly calculation mistakes ::cry2:


How did you get -0.08?
Original post by Chittesh14
How did you get that? This is what I would have done:

dy/dx = 5x^4 + 4x^3
d^2y/dx^2 = 20x^3 + 12x^2 = 0
x^2(20x + 12) = 0
x = 0 OR
x = -12/20 = - 6/10 = - 3/5


There is no stationary point at the point you said though.
Original post by Chittesh14
How did you get -0.08?


i got 0.05. silly mistakes again :cry:
Original post by Ano123
There is no stationary point at the point you said though.


-3/5 and 0 are the points of inflection for the curve given.
I'm sure not every point on inflection is a stationary point, or am I wrong?
A point of inflection is the turning point of the curve and that happens at -3/5 if you draw the graph out.
0 would be the stationary point in this case.
(edited 7 years ago)
Original post by Darshan98
i got 0.05. silly mistakes again :cry:


How did you get that lol?
Isn't it just what I showed in my working out as in 0 and -3/5.
Original post by Chittesh14
How did you get that lol?
Isn't it just what I showed in my working out as in 0 and -3/5.


no thats one of my points - (x,y) = (-0.6,0.05) and (0,0). The two points of inflection
Original post by Darshan98
no thats one of my points - (x,y) = (-0.6,0.05) and (0,0). The two points of inflection


Oh loool sorry. I'm an idiot, I didn't actually work out the coordinates of the points of inflection :s-smilie:.
Original post by Chittesh14
Oh loool sorry. I'm an idiot, I didn't actually work out the coordinates of the points of inflection :s-smilie:.


:lol: thats how i feel most of the times.
Original post by Chittesh14
-3/5 and 0 are the points of inflection for the curve given.


Hmmm, no inflection point at (0,0).
Try this question. I need help
Find the stationary points and determine their nature of the curve
y=0.1x5+x2-x3-x
Original post by Darshan98
:lol: thats how i feel most of the times.


Lol, you seem quite intelligent.
Have you finished the whole syllabus for the exam (FM GCSE)?
Original post by Ano123
Hmmm, no inflection point at (0,0).
Try this question. I need help
Find the stationary points and determine their nature of the curve
y=0.1x5+x2-x3-x


Well yeah, there may be no turning point at 0,0 I can agree. But, think of it as the stationary point then lol.
Original post by Chittesh14
Well yeah, there may be no turning point at 0,0 I can agree. But, think of it as the stationary point then lol.

There is a turning point at (0,0), not an inflection point though.

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