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 $\left(\frac{1}{3}, \frac{32}{27}\
ight)$
$f(x) = -x^3 - 4x^2 + 3x$
select all ordered pairs in the table which are located where the graph of $f(x)$ is decreasing.
ordered pairs
(-1 , -6) (2 , -18) (0 , 0)
(1 , -2) (-3 , -18) (-4 , -12)

Explanation:

Step1: Find derivative of $f(x)$

$f'(x) = \frac{d}{dx}(-x^3 -4x^2 +3x) = -3x^2 -8x +3$

Step2: Find critical points

Set $f'(x)=0$:
$-3x^2 -8x +3=0$
$3x^2 +8x -3=0$
Factor: $(3x-1)(x+3)=0$
Solutions: $x=-3$ and $x=\frac{1}{3}$

Step3: Test intervals for decrease

A function is decreasing when $f'(x)<0$.

• Interval $(-\infty, -3)$: Test $x=-4$, $f'(-4)=-3(-4)^2-8(-4)+3=-48+32+3=-13<0$ → decreasing
• Interval $(-3, \frac{1}{3})$: Test $x=0$, $f'(0)=0-0+3=3>0$ → increasing
• Interval $(\frac{1}{3}, \infty)$: Test $x=1$, $f'(1)=-3-8+3=-8<0$ → decreasing

Step4: Match pairs to intervals

• $(-4,-12)$: $x=-4 \in (-\infty, -3)$ → decreasing
• $(1,-2)$: $x=1 \in (\frac{1}{3}, \infty)$ → decreasing
• $(2,-18)$: $x=2 \in (\frac{1}{3}, \infty)$ → decreasing

Answer:

$(-4, -12)$, $(1, -2)$, $(2, -18)$