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Math Help - [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; July 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 f(x) and not 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?

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

    We know that 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 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 f is increasing for x\in(1,\;2)\cup(3,\;4) and decreasing for 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, f' must be positive on (1,\;2)\cup(3,\;4), and negative on (2,\;3).

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

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

    Now see which ones work:

    Quote Originally Posted by chukie View Post
    1)

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

    Quote Originally Posted by chukie View Post
    2)

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

    Quote Originally Posted by chukie View Post
    3)

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

    Quote Originally Posted by chukie View Post
    4)

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

    Quote Originally Posted by chukie View Post
    5)

    \begin{tabular}{|l|r|r|r|}<br />
\hline<br />
$x$: &      1.5 &    2.5 &   3.5\\\hline<br />
$f'(x)$: & -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, 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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    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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