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Thread: Cubic Function

  1. #1
    Super Member harpazo's Avatar
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    Cubic Function

    Use demos or a graphing calculator to graph the given function over the given interval.

    Do the following:

    A. Approximate any local max values and local min values.
    B. Determine where the function is increasing and decreasing.
    C. Round answers to two decimal places.

    I need the steps for A and B above.

    Here is the cubic function:

    f(x) = x^3 - 3x^2 + 5

    Given interval is (-1, 3).
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  2. #2
    Member Walagaster's Avatar
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    Re: Cubic Function

    What don't you understand about "use demos or a graphing calculator"? Google tells me that demos calculator is at https://www.desmos.com/calculator.
    Thanks from DenisB and topsquark
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  3. #3
    Super Member harpazo's Avatar
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    Re: Cubic Function

    Quote Originally Posted by Walagaster View Post
    What don't you understand about "use demos or a graphing calculator"? Google tells me that demos calculator is at https://www.desmos.com/calculator.
    How is this done without demos? How is this done by hand?
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  4. #4
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    Re: Cubic Function

    Quote Originally Posted by harpazo View Post
    How is this done without demos? How is this done by hand?
    Desmos is a graphing program to use if you don't have a graphics calculator - google it and download.

    You can draw the graph if you use calculus, but I don't think you are up to that yet. The only alternative at this stage is to plot a lot of points and join them up.
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  5. #5
    Super Member harpazo's Avatar
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    Re: Cubic Function

    Thank you.
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Cubic Function

    Consider that:

    $\displaystyle (x+1)^3=x^3+3x^2+3x+1$

    $\displaystyle -3(x+1)^2=-3x^2-6x-3$

    Adding the two expressions we get:

    $\displaystyle x^3-3x-2$

    And so:

    $\displaystyle g(x)=f(x+1)-3=x(x^2-3)$

    We can then see that $\displaystyle g(x)$ must be:

    Increasing on: $\displaystyle (-\infty,-a)$

    Decreasing on: $\displaystyle (-a,a)$

    Increasing on: $\displaystyle (a,\infty)$

    Where $\displaystyle 0<a<\sqrt{3}$

    We can further deduce that $\displaystyle g(x)$ has a local maximum at:

    $\displaystyle (-a,g(-a))=(-a,-g(a))$

    And a local minimum at:

    $\displaystyle (a,g(a))$

    Now, suppose in our attempt to find $\displaystyle a$, we try:

    $\displaystyle a=1$

    As this is the only integral value allowed. We find:

    $\displaystyle g(1)=-2$

    Now, we can look at (where $\displaystyle 0<h<\sqrt{3}-1$):

    $\displaystyle g(1+h)=(1+h)^3-3(1+h)=1+3h+3h^2+h^3-3-3h=-2+3h^2+h^3>-2$

    Likewise:

    $\displaystyle g(1-h)=(1-h)^3-3(1-h)=1-3h+3h^2-h^3-3+3h=-2+3h^2-h^3>-2$

    We really "lucked out" with our choice of $\displaystyle a$.

    And so we can conclude that $\displaystyle g(x)$ has a local maximum at $\displaystyle (-1,2)$ and a local minimum at $\displaystyle (1,-2)$. Transforming these coordinates back to $\displaystyle f(x)$ we then have:

    Local maximum at $\displaystyle (-1+1,2+3)=(0,5)$

    Local minimum at $\displaystyle (1+1,-2+3)=(2,1)$

    Increasing on $\displaystyle (-\infty,0)$

    Decreasing on $\displaystyle (0,2)$

    Increasing on $\displaystyle (2,\infty)$
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  7. #7
    Super Member harpazo's Avatar
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    Re: Cubic Function

    Quote Originally Posted by MarkFL View Post
    Consider that:

    $\displaystyle (x+1)^3=x^3+3x^2+3x+1$

    $\displaystyle -3(x+1)^2=-3x^2-6x-3$

    Adding the two expressions we get:

    $\displaystyle x^3-3x-2$

    And so:

    $\displaystyle g(x)=f(x+1)-3=x(x^2-3)$

    We can then see that $\displaystyle g(x)$ must be:

    Increasing on: $\displaystyle (-\infty,-a)$

    Decreasing on: $\displaystyle (-a,a)$

    Increasing on: $\displaystyle (a,\infty)$

    Where $\displaystyle 0<a<\sqrt{3}$

    We can further deduce that $\displaystyle g(x)$ has a local maximum at:

    $\displaystyle (-a,g(-a))=(-a,-g(a))$

    And a local minimum at:

    $\displaystyle (a,g(a))$

    Now, suppose in our attempt to find $\displaystyle a$, we try:

    $\displaystyle a=1$

    As this is the only integral value allowed. We find:

    $\displaystyle g(1)=-2$

    Now, we can look at (where $\displaystyle 0<h<\sqrt{3}-1$):

    $\displaystyle g(1+h)=(1+h)^3-3(1+h)=1+3h+3h^2+h^3-3-3h=-2+3h^2+h^3>-2$

    Likewise:

    $\displaystyle g(1-h)=(1-h)^3-3(1-h)=1-3h+3h^2-h^3-3+3h=-2+3h^2-h^3>-2$

    We really "lucked out" with our choice of $\displaystyle a$.

    And so we can conclude that $\displaystyle g(x)$ has a local maximum at $\displaystyle (-1,2)$ and a local minimum at $\displaystyle (1,-2)$. Transforming these coordinates back to $\displaystyle f(x)$ we then have:

    Local maximum at $\displaystyle (-1+1,2+3)=(0,5)$

    Local minimum at $\displaystyle (1+1,-2+3)=(2,1)$

    Increasing on $\displaystyle (-\infty,0)$

    Decreasing on $\displaystyle (0,2)$

    Increasing on $\displaystyle (2,\infty)$
    Wow! Very involved. Very mathematical. Definitely above my training.
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