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March 29, 2017

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For f(x)=2(x+5)^3 +7
Find and classify the extreme values, determine where the function is increasing and decreasing, where it is concave up and concave down, and any points of inflection.

I understand this is using the first and second derivative tests, but I can't find any zeroes for the derivative of this function. Am I doing something wrong?

  • Calculus - ,

    f(x)=0=2(x+5)^3 +7
    (x+5)^3=-7/2
    x+5= -cubroot(7/2)
    x=-5-cubroot(7/2)

    Those are the zeroes, but I don't see in the problem wording it asked for a zero

  • Calculus - ,

    y = 2 (x+5)^3 + 7

    to make this really easy you can say
    z = x+5
    then dz/dx = 1

    y = 2 z^3 + 7
    dy/dz = 6 z^2
    d^2y/dz^2 = 12 z

    so
    dy/dx = dy/dz * dz/dx = 6(x^2+10x+25)
    when is that zero ?
    (x+5)(x+5) = 0
    x = -5 twice

    what is d/dx (dy/dx)?
    6 ( 2 x + 10)
    when x = -5
    2(-5)+10 = 0
    so that is not a max or a min but an inflection point.

  • Calculus - ,

    Think of the graph for y = x^3
    It has no max/min, but just an inflection point at x=0.

    You function is just the same, only stretched by 2, shifted left by 5, and shifted up by 7.

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