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Core 2 - Section B help!

http://www.mei.org.uk/files/papers/c208ju_oplh.pdf
it is question 11, part 3 of this. I have differentiated y = x3 + 5x2 − 4x − 20 but I'm stumped with how to show the min point is 0.4 or how to answer this part of the question.
Any help is appreciated!:smile:
Original post by biggles98
http://www.mei.org.uk/files/papers/c208ju_oplh.pdf
it is question 11, part 3 of this. I have differentiated y = x3 + 5x2 − 4x − 20 but I'm stumped with how to show the min point is 0.4 or how to answer this part of the question.
Any help is appreciated!:smile:

Put that the differentiation is equal to zero. Because gradient is zero at turning points.
Reply 2
Avatar for biggles98
biggles98
OP
1
But what do I do after that?
Original post by biggles98
But what do I do after that?


Show it's a minimum by finding the second derivative and showing it's greater than zero.
So you find out value's of when f'(x) is zero
Then you substitute the values into F''(x)
If the answer is greater than zero; that's the co-ordinate for minimum
If it's less than zero, that's the coordinate for a maximum.
If F'''(X) isn't 0 then it's an Inflexion
(edited 9 years ago)
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davros
16
Original post by Chicken Fajita
Show it's a minimum by finding the second derivative and showing it's greater than zero.
So you find out value's of when f'(x) is zero
Then you substitute the values into F''(x)
If the answer is greater than zero; that's the co-ordinate for minimum
If it's less than zero, that's the coordinate for a maximum.
If F'''(X) is 0 then it's an Inflection


I presume you meant F''(x). And if it is equal to 0 it could still be a max or a min as well as an inflection - you need to check further!
Original post by davros
I presume you meant F''(x). And if it is equal to 0 it could still be a max or a min as well as an inflection - you need to check further!

No i meant the third derivative
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davros
16
Original post by Chicken Fajita
No i meant the third derivative


Well the same comment applies in general - the third derivative could still be 0 at a max or a min!

In this case they've given you a cubic that clearly has a max and a min and there aren't any other possibilities since dy/dx is quadratic and can only be 0 at 2 points so one is the max and the other is the min. :smile:
Original post by davros
Well the same comment applies in general - the third derivative could still be 0 at a max or a min!

In this case they've given you a cubic that clearly has a max and a min and there aren't any other possibilities since dy/dx is quadratic and can only be 0 at 2 points so one is the max and the other is the min. :smile:


I meant if it wasn't zero, sorry i typed it wrong
So if dy/dx = 0 it's turning point
If d2y/dx2<0 it's a max
If d2y/dx2>0 it's a min
If d3y/dx3 ​doesn't = 0 then it's a point of inflexion
I was just stating about the third derivative incase it came up just for extra information.
(edited 9 years ago)
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Avatar for davros
davros
16
Original post by Chicken Fajita
I meant if it wasn't zero, sorry i typed it wrong
So if dy/dx = 0 it's turning point
If d2y/dx2<0 it's a max
If d2y/dx2>0 it's a min
If d3y/dx3 ​doesn't = 0 then it's a point of inflexion
I was just stating about the third derivative incase it came up just for extra information.


Tbh I think they've pretty much stopped testing people on inflexions because people found it too confusing :smile:

For a C2 question I think you're just meant to look at the graph and say "two turning points, one is a max and one is a min, dy/dx = 0 will have 2 roots so which is which?" and that sorts it out (one root will be smaller than the other, so it's the leftmost turning point)
Reply 9
Avatar for ErwinJ
ErwinJ
10
You do not need to find the second derivative for this part of the question. The diagram shows one of the turning points has a negative x coordinate, whilst the other (the minimum point)has a positive x coordinate. Thus, when you get two solutions for your (dy/dx)=0 you need the positive x coordinate given.
(edited 9 years ago)

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