# Thread: [SOLVED] Critical values

1. ## [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

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?

2. 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:

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?

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

We know that $f'$ 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

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

Now see which ones work:

Originally Posted by chukie
1)

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

Originally Posted by chukie
2)

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

Originally Posted by chukie
3)

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

Originally Posted by chukie
4)

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

Originally Posted by chukie
5)

$\begin{tabular}{|l|r|r|r|}
\hline
x: & 1.5 & 2.5 & 3.5\\\hline
f'(x): & -2 & -4 & -2\\\hline\end{tabular}$
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, $f'$ must either be zero or undefined at the critical points.

3. Hmm yah I was confused too. I doubled check my problem, the first table lists values for f'(x). Maybe it was a typo?

4. Originally Posted by chukie
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.