Question

select the correct ordered pairs in the table. consider the function below, which has a relative minimum located at (-3, -18) and a relative maximum located at (\frac{1}{3}, \frac{14}{27}). f(x) = -x^3 - 4x^2 + 3x select all ordered pairs in the table which are located where the graph of f(x) is decreasing. \

$$\begin{tabular}{|c|c|c|} \\hline \\multicolumn{3}{|c|}{ordered pairs} \\\\ \\hline (-1, -6) & (2, -18) & (0, 0) \\\\ \\hline (1, -2) & (-3, -18) & (-4, -12) \\\\ \\hline \\end{tabular}$$

Explanation:

To determine where the function \( f(x) = -x^3 - 4x^2 + 3x \) is decreasing, we first find the critical points (where the derivative is zero or undefined) and then test intervals around these points.

Step 1: Find the derivative of \( f(x) \)

The derivative \( f'(x) \) of \( f(x) = -x^3 - 4x^2 + 3x \) is:
\[
f'(x) = -3x^2 - 8x + 3
\]

Step 2: Find critical points by solving \( f'(x) = 0 \)

We solve the quadratic equation \( -3x^2 - 8x + 3 = 0 \). Multiply both sides by -1:
\[
3x^2 + 8x - 3 = 0
\]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3 \), \( b = 8 \), and \( c = -3 \):
\[
x = \frac{-8 \pm \sqrt{64 + 36}}{6} = \frac{-8 \pm \sqrt{100}}{6} = \frac{-8 \pm 10}{6}
\]
This gives two solutions:
\[
x = \frac{-8 + 10}{6} = \frac{2}{6} = \frac{1}{3}
\]
\[
x = \frac{-8 - 10}{6} = \frac{-18}{6} = -3
\]
So the critical points are \( x = -3 \) and \( x = \frac{1}{3} \), which divide the number line into three intervals:

1. \( (-\infty, -3) \)
2. \( (-3, \frac{1}{3}) \)
3. \( (\frac{1}{3}, \infty) \)

Step 3: Test intervals for sign of \( f'(x) \)

• For \( (-\infty, -3) \), let's pick \( x = -4 \). Plug into \( f'(x) \):

\[
f'(-4) = -3(-4)^2 - 8(-4) + 3 = -48 + 32 + 3 = -13 < 0
\]
So \( f(x) \) is decreasing on \( (-\infty, -3) \).

• For \( (-3, \frac{1}{3}) \), let's pick \( x = 0 \). Plug into \( f'(x) \):

\[
f'(0) = -3(0)^2 - 8(0) + 3 = 3 > 0
\]
So \( f(x) \) is increasing on \( (-3, \frac{1}{3}) \).

• For \( (\frac{1}{3}, \infty) \), let's pick \( x = 2 \). Plug into \( f'(x) \):

\[
f'(2) = -3(2)^2 - 8(2) + 3 = -12 - 16 + 3 = -25 < 0
\]
So \( f(x) \) is decreasing on \( (\frac{1}{3}, \infty) \).

Step 4: Check each ordered pair

• \( (-1, -6) \): \( x = -1 \) is in \( (-3, \frac{1}{3}) \) (increasing interval), so not decreasing.
• \( (2, -18) \): \( x = 2 \) is in \( (\frac{1}{3}, \infty) \) (decreasing interval), so decreasing.
• \( (0, 0) \): \( x = 0 \) is in \( (-3, \frac{1}{3}) \) (increasing interval), so not decreasing.
• \( (1, -2) \): \( x = 1 \) is in \( (\frac{1}{3}, \infty) \) (decreasing interval), so decreasing.
• \( (-3, -18) \): This is a critical point (relative minimum), not part of a decreasing or increasing interval in the sense of the function's behavior around it (it's the minimum, so the function changes from decreasing to increasing here). But since we are looking for points where the graph is decreasing, and at \( x = -3 \), the function stops decreasing (starts increasing), we don't consider this as part of the decreasing interval (the interval before -3 is decreasing, but -3 is the endpoint).
• \( (-4, -12) \): \( x = -4 \) is in \( (-\infty, -3) \) (decreasing interval), so decreasing.

Answer:

The ordered pairs where the graph of \( f(x) \) is decreasing are \( (-4, -12) \), \( (1, -2) \), and \( (2, -18) \).