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:
Originally Posted by
chukie 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}{lrrrr}
\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}{lrrr}
\hline
$x$: & 1.5 & 2.5 & 3.5\\\hline
$f'(x)$: & +\; & \;\, & +\;\\\hline\end{tabular}$
Now see which ones work:
Originally Posted by
chukie 1)
$\displaystyle \begin{tabular}{lrrr}
\hline
$x$: & 1.5 & 2.5 & 3.5\\\hline
$f'(x)$: & 3 & 7 & 4\\\hline\end{tabular}$
This works!
Originally Posted by
chukie 2)
$\displaystyle \begin{tabular}{lrrr}
\hline
$x$: & 1.5 & 2.5 & 3.5\\\hline
$f'(x)$: & 2 & 4 & 2\\\hline\end{tabular}$
This doesn't!
Originally Posted by
chukie 3)
$\displaystyle \begin{tabular}{lrrr}
\hline
$x$: & 1.5 & 2.5 & 3.5\\\hline
$f'(x)$: & 2.5 & 8 & 3\\\hline\end{tabular}$
This doesn't!
Originally Posted by
chukie 4)
$\displaystyle \begin{tabular}{lrrr}
\hline
$x$: & 1.5 & 2.5 & 3.5\\\hline
$f'(x)$: & 2 & 4 & 2\\\hline\end{tabular}$
Nor does this.
Originally Posted by
chukie 5)
$\displaystyle \begin{tabular}{lrrr}
\hline
$x$: & 1.5 & 2.5 & 3.5\\\hline
$f'(x)$: & 2 & 4 & 2\\\hline\end{tabular}$
Nope.
Originally Posted by
chukie 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.