Just one sec...
Hey! Sign in to get help with your study questionsNew here? Join for free to post

How do find the minimum/maximum point on a curve? (C1 EDEXCEL)

Announcements Posted on
Take our short survey, £100 of Amazon vouchers to be won! 23-09-2016
    • Thread Starter
    Offline

    1
    ReputationRep:
    How do find the minimum/maximum point on a curve? and can you give an example please

    By the way keep this to Core 1 level/AS level

    Exam board is edexcel.

    Im a bit confused because the book doesn't mention anything about them, but sometimes I need to calculate them.

    Thanks
    Offline

    3
    ReputationRep:
    when you have a function, y=x^2 or something like that, you diffrentiate the function and make dy/dx=0, thats to find the stationary points, if you then diffrentiate again(this example is not useful for this bit) and you sub in your values obtained from dy/dx=0 then if the outcome number is above 0 then it is a minimum, if it is below 0 then it is a maximum. Look in the spoiler for my example
    Spoiler:
    Show
    y=x^3 - 2x, dy/dx = 3x^2 - 2, 3x^2 - 2 = 0, x^2 = 2/3 so x = plus or minus root 2/3. find d^2y/dx^2, this means differentiating 3x^2 - 2, this gives you 6x, sub the plus or minus root 2/3 in and work out which one is a maximum or a minimum
    EDIT: sorry its a bit messy, I tried to avoid latex because its hard to use and annoying at times
    Offline

    2
    ReputationRep:
    http://tutorial.math.lamar.edu/Class...calPoints.aspx
    http://tutorial.math.lamar.edu/Class...MaxValues.aspx

    Awesome site for other stuff too
    Offline

    2
    ReputationRep:
    (Original post by Eloades11)
    when you have a function, y=x^2 or something like that, you diffrentiate the function and make dy/dx=0, thats to find the stationary points, if you then diffrentiate again(this example is not useful for this bit) and you sub in your values obtained from dy/dx=0 then if the outcome number is above 0 then it is a minimum, if it is below 0 then it is a maximum. Look in the spoiler for my example
    Spoiler:
    Show
    y=x^3 - 2x, dy/dx = 3x^2 - 2, 3x^2 - 2 = 0, x^2 = 2/3 so x = plus or minus root 2/3. find d^2y/dx^2, this means differentiating 3x^2 - 2, this gives you 6x, sub the plus or minus root 2/3 in and work out which one is a maximum or a minimum
    EDIT: sorry its a bit messy, I tried to avoid latex because its hard to use and annoying at times
    What happens when you second differentiate for example: -6x+2
    you get -6
    you cannot sub any of the values obtained from first differentiation. what to do then?

    thanks!
    Offline

    3
    ReputationRep:
    (Original post by dream123)
    What happens when you second differentiate for example: -6x+2
    you get -6
    you cannot sub any of the values obtained from first differentiation. what to do then?

    thanks!
    y=-6x + 2
    dy/dx = -6
    when dy/dx=0
    -6=0?
    no stationary points, as you should already know if you just sketch the graph of -6x + 2
    (cant find maximum and minimum points if theres no stationary pionts!)
    hope this helped
    Offline

    3
    ReputationRep:
    (Original post by jayseanfan)
    How do find the minimum/maximum point on a curve? and can you give an example please

    By the way keep this to Core 1 level/AS level

    Exam board is edexcel.

    Im a bit confused because the book doesn't mention anything about them, but sometimes I need to calculate them.

    Thanks
    Use of differentiation to solve problems with Max/Min points is part of C2 not C1 for Edexcel

    You might be expected to use completed square to find the minimum point of a quadratic but that's all. [ the min of (x + p)^2 + q occurs at (-p,q) ]

    e.g. x^2 + 6x + 5
    = (x + 3)^2 - 4
    minimum is at (-3, -4)

    If you are asked to sketch a graph of a given function you will be required to show where it crosses the axes
    If you are asked to sketch a transformation of a graph you are reuired to label the new coordinates of points labelled on the original graph
    Offline

    3
    ReputationRep:
    (Original post by gdunne42)
    Use of differentiation to solve problems with Max/Min points is part of C2 not C1 for Edexcel
    If I knew this I would have pointed it out, but im on OCR doing A2 at the moment, similar topics though right? Plus people assumably doing C1 will do C2, but it would take a bit of pressure off if you learnt them in order. I ended up revising C4 for my C3 mock :/
    Offline

    2
    ReputationRep:
    (Original post by jayseanfan)
    How do find the minimum/maximum point on a curve? and can you give an example please

    By the way keep this to Core 1 level/AS level

    Exam board is edexcel.

    Im a bit confused because the book doesn't mention anything about them, but sometimes I need to calculate them.

    Thanks
    Wait, I'm doing Edexcel Maths, C1 and I've never been asked to calculate this. There's nothing in the book about it either. Can someone just confirm whether or not this could be in the exam, because I'm confused now, and we haven't been taught this :confused:
    Offline

    0
    ReputationRep:
    Nope thats C2.
    Offline

    3
    ReputationRep:
    (Original post by Eloades11)
    If I knew this I would have pointed it out, but im on OCR doing A2 at the moment, similar topics though right? Plus people assumably doing C1 will do C2, but it would take a bit of pressure off if you learnt them in order. I ended up revising C4 for my C3 mock :/
    Yep pretty much the same topics on the different awarding bodies but differences in which are in C1 and which in C2. Finding turning points/stationary points by setting dy/dx = 0 is C2 for Edexcel. Finding d^2y/dx^2 of a function is in Edexcel C1 and has occassionally been asked in the exam but you don't learn to do anything with it in terms of max/min points until C2. They will eventually cover it but don't need to know it for their January C1 exam.

    Specification: http://www.edexcel.com/migrationdocu...2%20180510.pdf
    Offline

    0
    ReputationRep:
    e.g.

    y = x^2 + 6x, find the coordinates of the minimum point

    start by differentiating (dy/dx will give you the formula for the gradient)

    dy/dx = 2x + 6

    minimum point => dy/dx = 0 (remember, at minimum/maximum points the gradient is 0 since at this particular instant the graph isn't increasing or decreasing)

    therefore 2x + 6 = 0
    2x = 6
    x = 3

    sub x = 3 into y = x^2 + 6x to get y = 27

    the coordinates of your minimum point are therefore (3,27)

    (main point: dy/dx will give you the formula for the gradient)
    Online

    2
    ReputationRep:
    now what if you where given a function like this y=x^3/x^2-1 and told to find max and min points of that curve?
    Offline

    3
    ReputationRep:
    (Original post by shady2.0)
    now what if you where given a function like this y=x^3/x^2-1 and told to find max and min points of that curve?
    Mate, this is a 5 year old thread...
    Online

    2
    ReputationRep:
    (Original post by Zacken)
    Mate, this is a 5 year old thread...
    yeah I understand mate am just new here
    Offline

    2
    ReputationRep:
    (Original post by shady2.0)
    now what if you where given a function like this y=x^3/x^2-1 and told to find max and min points of that curve?
    I assume you meant  y=\frac{x^3}{x^2 - 1} ?
    Online

    2
    ReputationRep:
    (Original post by Zacken)
    Mate, this is a 5 year old thread...
    so u gat the answers?
    Online

    2
    ReputationRep:
    (Original post by wmwaimeng)
    Did you mean  y=\frac{x^3}{x^2 - 1} or  y=\frac{x^3}{x^2} - 1 ?
    the first one
    Online

    2
    ReputationRep:
    (Original post by shady2.0)
    the first one
    u gat some answers for that?
    Offline

    2
    ReputationRep:
    (Original post by shady2.0)
    the first one
    Using the quotient rule, differentiate the equation and obtain dy/dx then equate it to 0, solve this equation.
    Spoiler:
    Show
     \frac{dy}{dx} = \frac{(x^2 - 1)(3x^2) - (x^3)(2x)}{(x^2 - 1)^2} = \frac{(x^2)(x^2 - 3)}{(x^2 - 1)^2} = 0

    Hence  (x^2)(x^2 - 3) = 0 and  x=0, \pm{\sqrt(3)}
    Then find the y-coordinates from the x values obtained.
    Online

    2
    ReputationRep:
    (Original post by wmwaimeng)
    Using the quotient rule, differentiate the equation and obtain dy/dx then equate it to 0, solve this equation.
    Spoiler:
    Show
     \frac{dy}{dx} = \frac{(x^2 - 1)(3x^2) - (x^3)(2x)}{(x^2 - 1)^2} = \frac{(x^2)(x^2 - 3)}{(x^2 - 1)^2} = 0

    Hence  (x^2)(x^2 - 3) = 0 and  x=0, \pm{\sqrt(3)}
    Then find the y-coordinates from the x values obtained.
    ohkay thank you buddy ..you are a life saver
 
 
 
Updated: November 7, 2015
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Poll
Who will be the next permanent England boss?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.