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Math Help - Calculus Question: Relative Max/Min and Inflection Points?

  1. #1
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    Calculus Question: Relative Max/Min and Inflection Points?

    For the function y = f(x), the solutions to the equation 0 = f'(x) are x = -3,0,2 and the solutions to the equation 0 - f"(x) are x= -1, 2. Furthermore, f"(-2) = 7, f"(1) = -9, and f"(3) = 4.

    a. Where does y = f(x) have a relative maximum and where does y = f(x) have a relative minimum?
    b. Where does y= f(x) have an inflection point?

    Would love some help with this, I don't even know where to begin!
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    Re: Calculus Question: Relative Max/Min and Inflection Points?

    The sign of the f'(x_0) tells you if the function f is increasing or decreasing in the neighborhood of x_0. Note that on left of the relative maximum f must be an increasing function, and on the right of the relative maximum it must be a decreasing function. It is the other way around in the neighborhood of a relative minimum.

    And bear in mind that what f' tells you about f, f'' tells you about f'.

    Solution to the f''(x)=0 is the candidate for the inflection point, and it actually is an inflection point if on the left and on the right of it f'' is of the opposite sign.
    Last edited by MathoMan; April 3rd 2012 at 02:37 PM.
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    Re: Calculus Question: Relative Max/Min and Inflection Points?

    So how do I go about answering this? Where is the starting point?

    Thanks for your reply.
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    Re: Calculus Question: Relative Max/Min and Inflection Points?

    First, x=-1 and x=2 are inflection points for function f since f'' is of the opposite sign on the opposite sides of either of those two numbers.
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    Re: Calculus Question: Relative Max/Min and Inflection Points?

    So to answer B, the inflection points are x=-1 and x=2? I do not understand how to get relative maximum and minimum since an equation was never given (or I don't know how to get an equation from the derivative values). My instructor has never presented us with a problem like this. I would appreciate being walked through it step-by-step since I have an exam over it tomorrow. If that would be at all possible, I would appreciate it.
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    Re: Calculus Question: Relative Max/Min and Inflection Points?

    Those inflection points split up the domain (suppose \mathb{R}) into intervals: \langle-\infty, -1\rangle, \langle -1,2\rangle, \langle 2, +\infty \rangle. On each of those intervals you know the sign of f'' since it can change sign only in inflection points.

    If you know that f''(x)>0, \forall x\in \langle a,b \rangle than f is convex on  \langle a,b \rangle. Otherwise f is said to be concave.

    Since you know that f''(-2)=7>0 and -2\in \langle-\infty, -1\rangle then f''(x)>0, \forall x\in\langle-\infty, -1\rangle, therefore f is convex on \langle-\infty, -1\rangle. Therefore the stationary point x=-3\in \langle -\infty,-1\rangle can only be relative minimum.

    Since you know that f''(1)=-9<0 and 1\in\langle -1,2\rangle then f''(x)<0, \forall x\in\langle -1,2\rangle, therefore f is concave on \langle -1,2\rangle. Therefore the stationary point x=0\in\langle -1,2\rangle can only be relative maximum.
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    Re: Calculus Question: Relative Max/Min and Inflection Points?

    Do the last one yourself.
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