12th Calculus

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use the derivative of the function
y=f(x)to find the points at which f has a

local maximum
local minimum
point of inflection

y'=(x-1)^2(x-2)(x-4)

  • 12th Calculus -

    if f(x)=uvw
    then
    f'(x)=uvw'+ uwv'+ wvu'

  • 12th Calculus -

    I will assume you really meant to write the derivative y' and not the function y although I would have thought the book would ask you to find that derivative.

    Wherever that derivative, y', is zero, the curve is horizontal
    This is twice at x = 1 and at x =2 and at x = 4

    Look at the second derivative to see what happens there.
    y'=(x-1)^2 (x^2-6x+8)
    so
    y" = (x-1)^2(2x-6)+2(x-1)(x^2-6x+8)
    now what is y"at x = 1?
    y"(1) = Zero, so that is an inflection point
    what is y" at x = 2?
    y"(2) = -2+0 = -2
    That is a maximum because y" is negative
    what is y"(4)?
    y"(4) = 9(2)+2*3(16-24+8)
    y"(4) = 18+0 = 18
    so it is Minimum at x = 4
    y"

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