Questions: The function r(x)=-2x^3-18x^2+6 has one local minimum and one local maximum. Graph the function on your calculator to estimate these local extrema and the intervals on which r is increasing and decreasing. When you are giving multiple intervals on which the function is increasing or decreasing, use a comma in between the intervals, NOT a union symbol. r(x) has a local maximum of at x= r(x) has a local minimum of at x= . r(x) is increasing on the interval(s): r(x) is decreasing on the interval(s):

The function r(x)=-2x^3-18x^2+6 has one local minimum and one local maximum. Graph the function on your calculator to estimate these local extrema and the intervals on which r is increasing and decreasing. When you are giving multiple intervals on which the function is increasing or decreasing, use a comma in between the intervals, NOT a union symbol. r(x) has a local maximum of at x= r(x) has a local minimum of at x= . r(x) is increasing on the interval(s): r(x) is decreasing on the interval(s):

Transcript text: The function $r(x)=-2 x^{3}-18 x^{2}+6$ has one local minimum and one local maximum. Graph the function on your calculator to estimate these local extrema and the intervals on which $r$ is increasing and decreasing. When you are giving multiple intervals on which the function is increasing or decreasing, use a comma in between the intervals, NOT a union symbol. $r(x)$ has a local maximum of $\square$ at $x=$ $\square$ $r(x)$ has a local minimum of $\square$ at $x=$ $\square$ . $r(x)$ is increasing on the interval(s): $\square$ $r(x)$ is decreasing on the interval(s): $\square$

Solution

Solution Steps

Step 1: Find the derivative of the function

To find the local extrema and intervals of increase and decrease, we first need to find the derivative of the function $ r(x) = -2x^3 - 18x^2 + 6 $.

\[ r'(x) = \frac{d}{dx}(-2x^3 - 18x^2 + 6) = -6x^2 - 36x \]

Step 2: Find critical points

Set the derivative equal to zero to find the critical points.

\[ -6x^2 - 36x = 0 \]

Factor the equation:

\[ -6x(x + 6) = 0 \]

The critical points are $ x = 0 $ and $ x = -6 $.

Step 3: Determine the nature of critical points

Use the second derivative test to determine whether each critical point is a local minimum or maximum.

\[ r''(x) = \frac{d}{dx}(-6x^2 - 36x) = -12x - 36 \]

Evaluate the second derivative at each critical point:

At $ x = 0 $: \[ r''(0) = -12(0) - 36 = -36 \quad (\text{negative, so local maximum}) \]
At $ x = -6 $: \[ r''(-6) = -12(-6) - 36 = 36 \quad (\text{positive, so local minimum}) \]

Step 4: Calculate the local extrema

Substitute the critical points back into the original function to find the local extrema.

Local maximum at $ x = 0 $: \[ r(0) = -2(0)^3 - 18(0)^2 + 6 = 6 \]
Local minimum at $ x = -6 $: \[ r(-6) = -2(-6)^3 - 18(-6)^2 + 6 = -432 - 648 + 6 = -1074 \]

Step 5: Determine intervals of increase and decrease

Analyze the sign of the first derivative around the critical points.

For $ x < -6 $, choose $ x = -7 $: \[ r'(-7) = -6(-7)^2 - 36(-7) = -294 + 252 = -42 \quad (\text{negative, decreasing}) \]
For $ -6 < x < 0 $, choose $ x = -3 $: \[ r'(-3) = -6(-3)^2 - 36(-3) = -54 + 108 = 54 \quad (\text{positive, increasing}) \]
For $ x > 0 $, choose $ x = 1 $: \[ r'(1) = -6(1)^2 - 36(1) = -6 - 36 = -42 \quad (\text{negative, decreasing}) \]

Final Answer

$ r(x) $ has a local maximum of $ 6 $ at $ x = 0 $.
$ r(x) $ has a local minimum of $ -1074 $ at $ x = -6 $.
$ r(x) $ is increasing on the interval(s): $ (-6, 0) $.
$ r(x) $ is decreasing on the interval(s): $ (-\infty, -6), (0, \infty) $.

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