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Thread: [SOLVED] Critical values

  1. #1
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    [SOLVED] Critical values

    If all the critical values of f(x) are shown below in the chart, which of the following could be values for f'(x) at x=1.5,2.5 and 3.5?
    x: 1, 2, 3, 4
    f'(x): 0.5, 3, -5, -3


    Answer choices:
    1)
    x: 1.5, 2.5, 3.5
    f'(x): 3, -7, 4

    2)
    x: 1.5, 2.5 , 3.5
    f'(x): 2, -4, -2

    3)
    x: 1.5, 2.5, 3.5
    f'(x): 2.5, -8, -3

    4)
    x: 1.5, 2.5, 3.5
    f'(x): -2, -4, 2

    5)
    x: 1.5, 2.5, 3.5
    f'(x): -2, -4, -2

    I am really confused by this question, because for the critical values shown in the chart, isnt f'(x) suppose to be zero at 1, 2, 3 and 4? Can someone help me?
    Last edited by chukie; Jul 2nd 2008 at 04:14 PM.
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  2. #2
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    Are you sure that the first table doesn't list values for $\displaystyle f(x)$ and not $\displaystyle f'(x)$? Otherwise the problem does not make sense, because the derivative must either be zero or undefined at a critical point.

    Assuming that this is the case:

    Quote Originally Posted by chukie View Post
    If all the critical values of f(x) are shown below in the chart, which of the following could be values for f'(x) at x=1.5,2.5 and 3.5?

    $\displaystyle \begin{tabular}{|l|r|r|r|r|}
    \hline
    $x$: & 1 & 2 & 3 & 4\\\hline
    $f(x)$: & 0.5 & 3 & -5 & -3\\\hline\end{tabular}$
    What do you know about critical points?

    We know that $\displaystyle f'$ can change signs only at a critical point and nowhere else (but just because there is a critical point does not mean the derivative must change signs). So $\displaystyle f$ can change from increasing to decreasing or from decreasing to increasing only at the critical points.

    Plotting the points above, we get something like
    Code:
      y
    
      ^
      |
      |
     3+     *
      |
     2+
      |
     1+
      |  *
    --+--+--+--+--+----> x
      |  1  2  3  4
    -1+
      |
    -2+
      |
    -3+           *
      |
    -4+
      |
    -5+        *
      |
    So $\displaystyle f$ is increasing for $\displaystyle x\in(1,\;2)\cup(3,\;4)$ and decreasing for $\displaystyle x\in(2,\;3)$ (remember it can't do both on the same interval, because the change can only happen at a critical point). Thus, $\displaystyle f'$ must be positive on $\displaystyle (1,\;2)\cup(3,\;4)$, and negative on $\displaystyle (2,\;3)$.

    So at $\displaystyle x = 1.5,\;2.5,\;3.5,$ our table should look something like

    $\displaystyle \begin{tabular}{|l|r|r|r|}
    \hline
    $x$: & 1.5 & 2.5 & 3.5\\\hline
    $f'(x)$: & +\; & --\;\, & +\;\\\hline\end{tabular}$

    Now see which ones work:

    Quote Originally Posted by chukie View Post
    1)

    $\displaystyle \begin{tabular}{|l|r|r|r|}
    \hline
    $x$: & 1.5 & 2.5 & 3.5\\\hline
    $f'(x)$: & 3 & -7 & 4\\\hline\end{tabular}$
    This works!

    Quote Originally Posted by chukie View Post
    2)

    $\displaystyle \begin{tabular}{|l|r|r|r|}
    \hline
    $x$: & 1.5 & 2.5 & 3.5\\\hline
    $f'(x)$: & 2 & -4 & -2\\\hline\end{tabular}$
    This doesn't!

    Quote Originally Posted by chukie View Post
    3)

    $\displaystyle \begin{tabular}{|l|r|r|r|}
    \hline
    $x$: & 1.5 & 2.5 & 3.5\\\hline
    $f'(x)$: & 2.5 & -8 & -3\\\hline\end{tabular}$
    This doesn't!

    Quote Originally Posted by chukie View Post
    4)

    $\displaystyle \begin{tabular}{|l|r|r|r|}
    \hline
    $x$: & 1.5 & 2.5 & 3.5\\\hline
    $f'(x)$: & -2 & -4 & 2\\\hline\end{tabular}$
    Nor does this.

    Quote Originally Posted by chukie View Post
    5)

    $\displaystyle \begin{tabular}{|l|r|r|r|}
    \hline
    $x$: & 1.5 & 2.5 & 3.5\\\hline
    $f'(x)$: & -2 & -4 & -2\\\hline\end{tabular}$
    Nope.

    Quote Originally Posted by chukie View Post
    I am really confused by this question, because for the critical values shown in the chart, isnt f'(x) suppose to be zero at 1, 2, 3 and 4?
    Well, $\displaystyle f'$ must either be zero or undefined at the critical points.
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  3. #3
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    Hmm yah I was confused too. I doubled check my problem, the first table lists values for f'(x). Maybe it was a typo?
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  4. #4
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    Quote Originally Posted by chukie View Post
    Hmm yah I was confused too. I doubled check my problem, the first table lists values for f'(x). Maybe it was a typo?
    Yes, I would guess that it is a typo, especially considering that my interpretation leads to only one answer being valid.
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