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#1

find the coordinates of any stationary points on each curve. By evaluating d^2y/dx^2 ( d squared Y over DX squared ) at each stationary point, determine whether it is a minimum or maximum point.

A) y=x^4 - 8x^2 - 2

B) y= x- 6 ^ 1/2
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1 month ago
#2
U work out the dy/dx of a point on each side and then work out whether it’s a u curve or n curve from that 0
1 month ago
#3
To find the points in the first place do dy/dx of the curve = to 0 and then that’s ur x coords then plug that in to work out y
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#4
Thx i understand that but dont know how to do dy/dx for b?
(Original post by Bobbyawsome)
To find the points in the first place do dy/dx of the curve = to 0 and then that’s ur x coords then plug that in to work out y
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1 month ago
#5
Chain rule
(Original post by Tigergirl)
Thx i understand that but dont know how to do dy/dx for b?
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1 month ago
#6
(Original post by Tigergirl)
Thx i understand that but dont know how to do dy/dx for b?
6^0.5 is just a number no x term the dy/dx should just be 1
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1 month ago
#7
chain rule i think if its supposed to be (x-6)^1/2
(Original post by Tigergirl)
Thx i understand that but dont know how to do dy/dx for b?
Last edited by nj1234; 1 month ago
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1 month ago
#8
Differentiate and set y' = 0 to get A: y' = 4x^3 - 16x = 4x(x+2)(x-2), giving x = 0, 2, -2
Differentiate again to find y''= 12x^2 - 16

x = 0 gives y'' = -16, < 0 so maximum
x = 2 gives y'' = 32, > 0 so minimum
x= -2 gives y'' = 32, > 0 so minimum

The curve is a w shape.
For the record, setting y'' = 0 gives the x-coordinates of the points of inflection.

As for B, . it's unclear whether you mean x - 6^0.5 or (x-6)^0.5

y = x - 6^0.5 is a straight line with y'=1, so no stationary points anyway
y = (x-6)^0.5 with y'= 0.5(x-6)^-0.5 which is always positive and never = 0, so it's is an increasing function (draw it on desmos), so also no stationary points.
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1 month ago
#9
Ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh hhhhhhhhhhhhhhhhh i dead< what is a good palce for revision and understandable< i literally failed
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1 month ago
#10
(Original post by Bobbyawsome)
U work out the dy/dx of a point on each side and then work out whether it’s a u curve or n curve from that Not if the question specifically says use the second derivative. You'd get 0 marks if you did it your way I'm afraid.
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1 month ago
#11
(Original post by dextrous63)
Differentiate and set y' = 0 to get A: y' = 4x^3 - 16x = 4x(x+2)(x-2), giving x = 0, 2, -2
Differentiate again to find y''= 12x^2 - 16

x = 0 gives y'' = -16, < 0 so maximum
x = 2 gives y'' = 32, > 0 so minimum
x= -2 gives y'' = 32, > 0 so minimum

The curve is a w shape.
For the record, setting y'' = 0 gives the x-coordinates of the points of inflection.

As for B, . it's unclear whether you mean x - 6^0.5 or (x-6)^0.5

y = x - 6^0.5 is a straight line with y'=1, so no stationary points anyway
y = (x-6)^0.5 with y'= 0.5(x-6)^-0.5 which is always positive and never = 0, so it's is an increasing function (draw it on desmos), so also no stationary points.
Oh lol forgot the brackets I was gunna say it didn’t make sense if the gradient was just 1 too easy 😂
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1 month ago
#12
(Original post by dextrous63)
Not if the question specifically says use the second derivative. You'd get 0 marks if you did it your way I'm afraid.
It doesn’t say use the 2nd derivative tho... or r u on about a specific Q from an exam?? If so u did a different one to me
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1 month ago
#13
(Original post by Bobbyawsome)
It doesn’t say use the 2nd derivative tho... or r u on about a specific Q from an exam?? If so u did a different one to me
Er, "By evaluating d^2y/dx^2 ( d squared Y over DX squared ) at each stationary point, determine whether it is a minimum or maximum point."
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1 month ago
#14
(Original post by Russellster)
... what is a good palce for revision and understandable< i literally failed
Use examsolutions.net
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1 month ago
#15
(Original post by dextrous63)
Er, "By evaluating d^2y/dx^2 ( d squared Y over DX squared ) at each stationary point, determine whether it is a minimum or maximum point."
U can identify by taking the dy/dx of each side and if then by working with that u can tell whether the curve was shaped like an n or a u
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1 month ago
#16
(Original post by Bobbyawsome)
U can identify by taking the dy/dx of each side and if then by working with that u can tell whether the curve was shaped like an n or a u
Indeed you can, and this is particularly useful to determine whether a function is increasing or decreasing around a point of inflection. It's also the only realistic way of doing it if differentiating y' would be somewhat onerous!!!

However, if an exam question told you specifically to use the 2nd derivative (as the questions in the op do), then you'd have to do so if you wanted any marks.

Don't shoot the messenger Last edited by dextrous63; 1 month ago
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1 month ago
#17
(Original post by dextrous63)
Indeed you can, and this is particularly useful to determine whether a function is increasing or decreasing around a point of inflection. It's also the only realistic way of doing it if differentiating y' would be somewhat onerous!!!

However, if an exam question told you specifically to use the 2nd derivative (as the questions in the op do), then you'd have to do so if you wanted any marks.

Don't shoot the messenger Ok np I just do whatever I see in the question the easiest way possible lol maybe I didn’t quite understand it properly due to it being payed out on a screen style. and I’m only gcse so... I’d probably get away with it 😝
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#18
thanks

for b the question is y= x-6x^1/2 ( as in 6x with the half power 6x^1/2) I realised I have written it wrong in the original post
(Original post by dextrous63)
Differentiate and set y' = 0 to get A: y' = 4x^3 - 16x = 4x(x+2)(x-2), giving x = 0, 2, -2
Differentiate again to find y''= 12x^2 - 16

x = 0 gives y'' = -16, < 0 so maximum
x = 2 gives y'' = 32, > 0 so minimum
x= -2 gives y'' = 32, > 0 so minimum

The curve is a w shape.
For the record, setting y'' = 0 gives the x-coordinates of the points of inflection.

As for B, . it's unclear whether you mean x - 6^0.5 or (x-6)^0.5

y = x - 6^0.5 is a straight line with y'=1, so no stationary points anyway
y = (x-6)^0.5 with y'= 0.5(x-6)^-0.5 which is always positive and never = 0, so it's is an increasing function (draw it on desmos), so also no stationary points.
Last edited by Tigergirl; 1 month ago
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1 month ago
#19
(Original post by Bobbyawsome)
Ok np I just do whatever I see in the question the easiest way possible lol maybe I didn’t quite understand it properly due to it being payed out on a screen style. and I’m only gcse so... I’d probably get away with it 😝
Fair enough. I'm a maths teacher so have a bit of an advantage 0
1 month ago
#20
(Original post by dextrous63)
Fair enough. I'm a maths teacher so have a bit of an advantage Hahah ok 👍 although as a math teacher you’ll be happy to know the gcse today was perfect 👌
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