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Maximum and minimum value of a curve.

  • 28-09-2015 8:06pm
    #1
    Hedgecutter
    Registered Users, Registered Users 2 Posts: 1,054 ✭✭✭


    Wondering if I'm on the right track here.


Comments

  • #2
    MathsManiac
    Registered Users, Registered Users 2 Posts: 1,595 ✭✭✭MathsManiac


    Your differentiation is correct.

    However, to find the max and min points, you need to set the FIRST derivative equal to zero, not the second.

    If you do this, you should get two values of t (t = -1 and +1, actually).

    You use the second derivative to tell you which of these is a max and which is a min. (Put t=-1 into the second derivative and you get a negative number, telling you that -1 gives a max. Put t=1 into the second derivative and you get a positive number, telling you that +1 gives a min.)

    The y-values you get by substituting these values of t into the original equation are the local max and min values of the function.

    (By the way, when the independent variable is t rather than x, you should call the derivative dy/dt rather than dy/dx, and similarly for the 2nd derivative.)


  • #3
    Hedgecutter
    Registered Users, Registered Users 2 Posts: 1,054 ✭✭✭Hedgecutter


    MathsManiac wrote: »
    Your differentiation is correct.

    However, to find the max and min points, you need to set the FIRST derivative equal to zero, not the second.

    If you do this, you should get two values of t (t = -1 and +1, actually).

    You use the second derivative to tell you which of these is a max and which is a min. (Put t=-1 into the second derivative and you get a negative number, telling you that -1 gives a max. Put t=1 into the second derivative and you get a positive number, telling you that +1 gives a min.)

    The y-values you get by substituting these values of t into the original equation are the local max and min values of the function.


    (By the way, when the independent variable is t rather than x, you should call the derivative dy/dt rather than dy/dx, and similarly for the 2nd derivative.)


    Thanks


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