I understand that but the question asked for the point at which the gradient takes its maximum value. I assumed this was asking for the value of x at which the curve was the steepest.(Original post by AlfieH)
It's not the point at which the gradient is the maximum  so you may be getting confused there.
Nor is it the point where the gradient is the lowest.
Instead the gradient is always 0 at the turning points/ min or max points. It is the maximum/ minimum as in the highest/lowest POINT of the curve not gradient.
If that makes sense?
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OCR C3 (not MEI) Official Thread  Tuesday 21st June 2016
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 20062016 23:14
Post rating:1 
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 20062016 23:20
(Original post by SI 1)
I understand that but the question asked for the point at which the gradient takes its maximum value. I assumed this was asking for the value of x at which the curve was the steepest. 
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 63
 20062016 23:27
(Original post by Yazmin123)
can someone help me with this question from jan 20063sin6Bcosec2B=4 where beta lies between zero and 90 degreesany help would be appreciated 
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 64
 20062016 23:29
(Original post by AlfieH)
No  that means when the curve has a minimum point so to find x you must equate it to 0.
Basically when it asks for the maximum (or minimum) gradient, I make the second derivative equal to 0, solve for x and sub that value of x back into the first derivative. is that right? 
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 65
 20062016 23:33
(Original post by SI 1)
I understand that but the question asked for the point at which the gradient takes its maximum value. I assumed this was asking for the value of x at which the curve was the steepest.
Post the question if you are still unsure. 
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 66
 20062016 23:36
(Original post by 16characterlimit)
If you want to find the point at which the gradient is maximum you'll have to differentiate again and set f''(x)=0, then verify if this point is a maximum or minimum from looking at the graph of dy/dx or finding the third derivative and checking if its positive or negative (yeesh).
Post the question if you are still unsure. 
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 20062016 23:45
(Original post by SI 1)
its question 8 ii on this paper
d^2y/dx^2 = (22lnx)/x^2 found using product/quotient rule, I will refer to it as f''(x) from now cause its less messy
when f''(x) = 0 f'(x) is maximum, which is the gradient, hence 22lnx = 0 (times both sides by x^2) and rearrange to get lnx = 1 . x = e therefore.
Subbing back into f'(x) we find the maximum gradient is 2/e
Because we know the gradient, we can now can find out the tangent. We know at point P x=e therefore y=1, so that gives us enough information to write the tangent as y=2x/e  1
Which does indeed go through (0,1) 
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 68
 20062016 23:53
(Original post by 16characterlimit)
dy/dx = (2lnx)/x found using the chain rule, u=lnx
d^2y/dx^2 = (22lnx)/x^2 found using product/quotient rule, I will refer to it as f''(x) from now cause its less messy
when f''(x) = 0 f'(x) is maximum, which is the gradient, hence 22lnx = 0 (times both sides by x^2) and rearrange to get lnx = 1 . x = e therefore.
Subbing back into f'(x) we find the maximum gradient is 2/e
Because we know the gradient, we can now can find out the tangent. We know at point P x=e therefore y=1, so that gives us enough information to write the tangent as y=2x/e  1
Which does indeed go through (0,1) 
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 69
 20062016 23:57
(Original post by SI 1)
Thank you so much!! One question, is the fact that f''(x)=0 is when f'(x) is maximum a rule or something cos I don't remember ever learning that? 
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 20062016 23:59
(Original post by 16characterlimit)
Try imagining f'(x) as a different function g(x), if you wanted to find the maximum of g(x) what would you do? 
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 21062016 00:00
(Original post by 16characterlimit)
Try imagining f'(x) as a different function g(x), if you wanted to find the maximum of g(x) what would you do?
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 21062016 00:03
(Original post by SI 1)
Do you mean the maximum gradient or the maximum point? 
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 21062016 00:04
(Original post by drandy76)
From what I recall the second derivative being 0 at a point isn't informative of whether it's a minimum or maximum point?
Posted from TSR Mobile
edit: nvm I didnt read your post properly.
I think if its d^2y/dx^2 = 0 its an inflection point and you have to look at surrounding values of the graph, this very very rarely comes up in C1 and even less so in C3 thoughLast edited by 16characterlimit; 21062016 at 00:05. 
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 21062016 00:08
(Original post by 16characterlimit)
Point 
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 21062016 00:10
(Original post by 16characterlimit)
No, you should always check, but in the context of that question there is only one stationary point and it more or less told you you have to find a maximum.
edit: nvm I didnt read your post properly.
I think if its d^2y/dx^2 = 0 its an inflection point and you have to look at surrounding values of the graph, this very very rarely comes up in C1 and even less so in C3 though
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 21062016 00:12
(Original post by SI 1)
wouldn't you just find the stationary points and the put into second derivative to see which one is a maximum
e.g
f'(x)=0 gives min/max of f(x)
f''(x)=0 gives min/man of f'(x)
f^n(x)=0 gives min/max of f^n1(x) where n is the order of the derivative 
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 21062016 00:14
(Original post by 16characterlimit)
Yes, so basically the stationary points of a function are the values of x of the functions integral are maxima or minima.
e.g
f'(x)=0 gives min/max of f(x)
f''(x)=0 gives min/man of f'(x)
f^n(x)=0 gives min/max of f^n1(x) where n is the order of the derivative
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 21062016 00:15
(Original post by 16characterlimit)
Yes, so basically the stationary points of a function are the values of x of the functions integral are maxima or minima.
e.g
f'(x)=0 gives min/max of f(x)
f''(x)=0 gives min/man of f'(x)
f^n(x)=0 gives min/max of f^n1(x) where n is the order of the derivative 
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 79
 21062016 06:58
(Original post by mrbeady9)

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 21062016 07:41
(Original post by underestimate)
How did you change it into the first form, like what formula did you use?
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Updated: August 21, 2016
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