Turn on thread page Beta
    • Thread Starter
    Offline

    1
    ReputationRep:
    how to find the minimum points of y=x^4-8^2+3 ?
    do i need to differentiate it first?

    many thanks!
    • TSR Support Team
    Offline

    21
    ReputationRep:
    TSR Support Team
    (Original post by jianwen)
    how to find the minimum points of y=x^4-8^2+3 ?
    do i need to differentiate it first?

    many thanks!
    You need to use the fact that \frac{dy}{dx}=0 at minima.
    Offline

    15
    ReputationRep:
    You do need to differentiate the equation as this gives us a gradient function. So at a minimum point the gradient is 0. So you differentiate the equation and set it equal to 0. You then solve by factorising the cubic equation. And solvin for x. To see if it is a maximum point or minimum point (or inflection point) you then take the second derivative of the equation - by differentiating the derivative. Then plug in the x values you got for dy/dx = 0 and if the second derivative is greater than 0 then it is a min point at that x value. If the second derivative is less than 0 then it is a max point at that x value. If the second derivative is 0 then you have a point of inflection.
    • Thread Starter
    Offline

    1
    ReputationRep:
    thank you very much!
    (Original post by B_9710)
    You do need to differentiate the equation as this gives us a gradient function. So at a minimum point the gradient is 0. So you differentiate the equation and set it equal to 0. You then solve by factorising the cubic equation. And solvin for x. To see if it is a maximum point or minimum point (or inflection point) you then take the second derivative of the equation - by differentiating the derivative. Then plug in the x values you got for dy/dx = 0 and if the second derivative is greater than 0 then it is a min point at that x value. If the second derivative is less than 0 then it is a max point at that x value. If the second derivative is 0 then you have a point of inflection.
    • Thread Starter
    Offline

    1
    ReputationRep:
    thank you!!
    (Original post by Plagioclase)
    You need to use the fact that \frac{dy}{dx}=0 at minima.
    Online

    18
    ReputationRep:
    (Original post by jianwen)
    how to find the minimum points of y=x^4-8^2+3 ?
    do i need to differentiate it first?

    many thanks!
    (Original post by Plagioclase)
    You need to use the fact that \frac{dy}{dx}=0 at minima.
    (Original post by B_9710)
    You do need to differentiate the equation as this gives us a gradient function. So at a minimum point the gradient is 0. So you differentiate the equation and set it equal to 0. You then solve by factorising the cubic equation. And solvin for x. To see if it is a maximum point or minimum point (or inflection point) you then take the second derivative of the equation - by differentiating the derivative. Then plug in the x values you got for dy/dx = 0 and if the second derivative is greater than 0 then it is a min point at that x value. If the second derivative is less than 0 then it is a max point at that x value. If the second derivative is 0 then you have a point of inflection.
    For this question, you most definitely do NOT have to differentiate, and I would suspect you're not supposed to.

    Firstly, if you really mean y=x^4 -8^2 +3 I would hope that it is obvious this is minimized when x = 0.

    If (as seems likely) you mean y=x^4 - 8x^2 + 3, then treat it as a quadratic in x^2 and complete the square: y = (x^2 - 4)^2 -13, which is obviously mimimized when x^2 = 4, i.e. x = +/- 2.

    No need to differentiate, verifiy that the points where dy/dx = 0 are minimums as opposed to maximums etc.

    Edit: fixed stupid calculation error
    Offline

    4
    ReputationRep:
    (Original post by DFranklin)
    For this question, you most definitely do NOT have to differentiate, and I would suspect you're not supposed to.

    Firstly, if you really mean y=x^4 -8^2 +3 I would hope that it is obvious this is minimized when x = 0.

    If (as seems likely) you mean y=x^4 - 8x^2 + 3, then treat it as a quadratic in x^2 and complete the square: y = (x^2 - 4)^2 -13, which is obviously mimimized when x^2 = 2, i.e. x = +/- sqrt(2).
    .
    Do you not mean whenever x^2 = 4, i.e. x = +/- sqrt(4)?
    Online

    18
    ReputationRep:
    (Original post by Jordan\)
    Do you not mean whenever x^2 = 4, i.e. x = +/- sqrt(4)?
    Yes, thanks. (I checked x=2 as a minimum mentally and obviously still had 2 in my head when I wrote the next bit).
    Offline

    7
    ReputationRep:
    (Original post by B_9710)
    If the second derivative is 0 then you have a point of inflection.
    Not necessarily. I always mean to write this up systematically.

    1) For a "standard" x-y graph, If dy/dx = 0 then the gradient at that value of x is horizontal which you identify as a maximum , minimum or point of inflection. More accurately, the graph could be horizontal everywhere e.g. y = 2. For this function, first and second derivative are zero for all x but no P.O.I anywhere. A point of inflection can be sloping e.g. tan x for x = 0. Another example, between the maximum and minimum of a simple cubic, y = (x+1)(x-1)(x+2), the (sloping) point of inflection occurs where the derivative has a maximum or minimum (that's where the second derivative comes in).

    For y = x^3 at x= 0 then dy/dx=0 and d^2y/dx^2 = 0 and we do have P.O.I at x=0.

    Problem occurs for y= x^4. Here dy/dx and d^2y/dx^2 are zero for x=0 but NOT a P.O.I.

    I think an argument goes like this.

    Consider y = x^2(x+a)^2 . For a = 0 the first derivative has an odd number of coincident (repeated) roots at x= 0 , the second derivative is zero but there is NOT a P.O.I at x=0. For a \neq0 you can't create an even number of coincident roots for first derivative so can't create horizontal P.O.I for this function.

    Consider y = x^3(x+a) . For a \neq0 the first derivative has an even number of repeated roots at x = 0, and there is a horizontal P.O.I at x=0. As a approaches zero the derivative gets an odd number of repeated roots at x=0 and P.O.I becomes a minimum (no longer a P.O.I even though second derivative is zero here (at x = 0)).

    I am sure I can write that more clearly but I think the idea is (mostly) correct. I am sure someone will put me right if not. Whatever, this is why the textbooks avoid discussing nature of turning points when first and second derivative are zero for same value of x.

    The "old boys" were taught a rule of repeated differentiation (fourth, 5th, etc derivatives) to determine the "real nature" of every turning point and the location of sloping P.O.I.s.
    Offline

    14
    ReputationRep:
    (Original post by cliverlong)
    Not necessarily. I always mean to write this up systematically.

    1) For a "standard" x-y graph, If dy/dx = 0 then the gradient at that value of x is horizontal which you identify as a maximum , minimum or point of inflection. More accurately, the graph could be horizontal everywhere e.g. y = 2. A point of inflection can be sloping e.g. tan x for x = 0. Another example, between the maximum and minimum of a simple cubic, y = (x+1)(x-1)(x+2), the (sloping) point of inflection occurs where the derivative has a maximum or minimum (that's where the second derivative comes in).

    For y = x^3 at x= 0 then dy/dx=0 and d^2y/dx^2 = 0 and we do have P.O.I at x=0.

    Problem occurs for y= x^4. Here dy/dx and d^2y/dx^2 are zero for x=0 but NOT a P.O.I.

    I think an argument goes like this.

    Consider y = x^2(x+a)^2 . For a = 0 the first derivative has an odd number of coincident (repeated) roots at x= 0 , the second derivative is zero but there is NOT a P.O.I at x=0. For a \neq0 you can't create an even number of coincident roots for first derivative so can't create horizontal P.O.I for this function.

    Consider y = x^3(x+a) . For a \neq0 the first derivative has an even number of repeated roots at x = 0, and there is a horizontal P.O.I at x=0. As a approaches zero the derivative gets an odd number of repeated roots at x=0 and P.O.I becomes a minimum (no longer a P.O.I even though second derivative is zero here (at x = 0)).

    I am sure I can write that more clearly but I think the idea is (mostly) correct. I am sure someone will put me right if not. Whatever, this is why the textbooks avoid discussing nature of turning points when first and second derivative are zero for same value of x.

    The "old boys" were taught a rule of repeated differentiation (fourth, 5th, etc derivatives) to determine the "real nature" of every turning point and the location of sloping P.O.I.s.
    In a nutshell for higher powers you use the nth derivative test to check for local max, min and POI.
    Two conditions for a POI :
    1) f''(x) =0
    2) f'''(x) =\ 0.
    If both the 2nd and 3rd derivative =0 then use nth derivative test.
    Offline

    14
    ReputationRep:
    (Original post by DFranklin)
    For this question, you most definitely do NOT have to differentiate, and I would suspect you're not supposed to.

    Firstly, if you really mean y=x^4 -8^2 +3 I would hope that it is obvious this is minimized when x = 0.

    If (as seems likely) you mean y=x^4 - 8x^2 + 3, then treat it as a quadratic in x^2 and complete the square: y = (x^2 - 4)^2 -13, which is obviously mimimized when x^2 = 4, i.e. x = +/- 2.

    No need to differentiate, verifiy that the points where dy/dx = 0 are minimums as opposed to maximums etc.

    Edit: fixed stupid calculation error
    Wrong. You have to differentiate to find local max or local min points. This is maths 101.
    Online

    19
    ReputationRep:
    (Original post by stochasticking)
    Wrong. You have to differentiate to find local max or local min points. This is maths 101.
    what is maths 101?
    Offline

    7
    ReputationRep:
    (Original post by stochasticking)
    Wrong. You have to differentiate to find local max or local min points. This is maths 101.
    Look up use of completing the square of a quadratic function ("actual" or "hidden")to find turning points, my friend. You will then understand DF's method.

    clive
    Offline

    7
    ReputationRep:
    (Original post by stochasticking)
    In a nutshell for higher powers you use the nth derivative test to check for local max, min and POI.
    Two conditions for a POI :
    1) f''(x) =0
    2) f'''(x) =\ 0.
    If both the 2nd and 3rd derivative =0 then use nth derivative test.
    That's it ! Muchly thankly TeeEm.
    IS what I wrote correct (if rather verbose)? I need reassurance on this.
    Offline

    14
    ReputationRep:
    (Original post by jianwen)
    how to find the minimum points of y=x^4-8^2+3 ?
    do i need to differentiate it first?

    many thanks!
    Y= X^4-8x^2+3
    1) dy/dx= 4x^3-16x
    2) dy/dx = 4(x^3-4x)
    3) set dy/dx =0
    4) 0=4(x^3-4x)
    5) 0=x(x^2-4)
    6) x=0 and x= +\- 2 > critical values.
    7) d2y/dx^2= 12x^2-16
    8) evaluate at CV
    9) d^2y/dx^2= -16 at x=0 therefore local max at x=0
    10) d^2y/dx^2= 32 at x=+2 and x=-2. Therefore local minimum points at x=-2 and x=+2. Plug x values into original equation to get y values.
    You can set second differential to 0 and check for any POIs. If the second differential =0 and third differential does not equal zero you have a POI.
    Offline

    14
    ReputationRep:
    (Original post by cliverlong)
    That's it ! Muchly thankly TeeEm.
    IS what I wrote correct (if rather verbose)? I need reassurance on this.
    I didn't read all of it lol But the last paragraph you wrote is essentially it yeah
    Offline

    14
    ReputationRep:
    (Original post by cliverlong)
    Look up use of completing the square of a quadratic function ("actual" or "hidden"to find turning points, my friend. You will then understand DF's method.

    clive
    Hmm okay thank you, I will !
    Offline

    14
    ReputationRep:
    (Original post by stochasticking)
    Hmm okay thank you, I will !
    I got the same solutions as him , so now I look like a ******* lol hindsight is a beautiful thing :P
    Online

    18
    ReputationRep:
    (Original post by stochasticking)
    Wrong. You have to differentiate to find local max or local min points. This is maths 101.
    Fortunately, some of us have a maths education that didn't stop at maths 101.

    What you've written is errant nonsense, note that functions don't even have to be differentiable. So it would be tricky to use differentiation to find local max/mins for y = | x |.for example.

    Also, in terms of distinguishing min/max/inflexion:

    Define f(x) = exp(-1/x^2) (x non zero), f(0) = 0 and also
    g(x) = -exp(-1/x^2) (x non zero), g(0) = 0 and finally
    h(x) = exp(-1/x^2) (x > 0), -exp(-1/x^2) x < 0, h(0) = 0

    Have fun using differentiation to decide whether each of f, g or h have a minimum, maximum or point of inflexion at x = 0.

    Spoiler:
    Show
    f,g,h are all infinitely differentiable at x = 0, and all their derivatives are 0 there.
    Online

    19
    ReputationRep:
    Can someone tell me what maths 101 is?
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: September 11, 2015

University open days

  1. Norwich University of the Arts
    Postgraduate Open Days Postgraduate
    Thu, 19 Jul '18
  2. University of Sunderland
    Postgraduate Open Day Postgraduate
    Thu, 19 Jul '18
  3. Plymouth College of Art
    All MA Programmes Postgraduate
    Thu, 19 Jul '18
Poll
How are you feeling in the run-up to Results Day 2018?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

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

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

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