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Thread: Calculus Question: Relative Max/Min and Inflection Points?

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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 $\displaystyle f'(x_0)$ tells you if the function $\displaystyle f$ is increasing or decreasing in the neighborhood of $\displaystyle x_0$. Note that on left of the relative maximum $\displaystyle f$ 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 $\displaystyle f'$ tells you about $\displaystyle f$, $\displaystyle f''$ tells you about $\displaystyle f'$.

    Solution to the $\displaystyle f''(x)=0$ is the candidate for the inflection point, and it actually is an inflection point if on the left and on the right of it $\displaystyle f''$ is of the opposite sign.
    Last edited by MathoMan; Apr 3rd 2012 at 01:37 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, $\displaystyle x=-1$ and $\displaystyle x=2$ are inflection points for function $\displaystyle f$ since $\displaystyle f''$ 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 $\displaystyle \mathb{R}$) into intervals: $\displaystyle \langle-\infty, -1\rangle$,$\displaystyle \langle -1,2\rangle$,$\displaystyle \langle 2, +\infty \rangle$. On each of those intervals you know the sign of $\displaystyle f''$ since it can change sign only in inflection points.

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

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

    Since you know that $\displaystyle f''(1)=-9<0$ and $\displaystyle 1\in\langle -1,2\rangle$ then $\displaystyle f''(x)<0, \forall x\in\langle -1,2\rangle$, therefore $\displaystyle f$ is concave on$\displaystyle \langle -1,2\rangle$. Therefore the stationary point $\displaystyle x=0\in\langle -1,2\rangle$ 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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