Math / Calculus

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Determine where the function is (a) increasing; (b) decreasing; and (c) determine where relative extrema occur. Do not sketch the graph. $y=-\frac{x^{3}}{3}-2 x^{2}+12 x-3$ (a) For which interval(s) is the function increasing? Select the correct choice below and, if necessary, fill in the answer box within your choice. A. The function is increasing on $(-6,2)$ . (Type your answer in interval notation. Use a comma to separate answers as needed.) B. The function is never increasing. (b) For which interval(s) is the function decreasing? Select the correct choice below and, if necessary, fill in the answer box within your choice. A. The function is decreasing on $\square$ . (Type your answer in interval notation. Use a comma to separate answers as needed.) B. The function is never decreasing.

Studdy Solution

STEP 1

What is this asking? We need to find where the function $y = -\frac{x^3}{3} - 2x^2 + 12x - 3$ is going up, going down, and where it has its peaks and valleys. Watch out! Don't mix up increasing/decreasing with positive/negative!
A function can be decreasing but still have positive values.
Also, remember that extrema are points, not intervals.

STEP 2

1. Find the derivative.
2. Find critical points.
3. Analyze intervals.

STEP 3

To figure out where a function is increasing or decreasing, we need to look at its derivative, which tells us about the slope of the function.
A positive derivative means the function is going uphill (increasing), a negative derivative means it's going downhill (decreasing), and a zero derivative usually means we've found a peak or a valley!
Let's find the derivative of our function $y = -\frac{x^3}{3} - 2x^2 + 12x - 3$ .

STEP 4

Using the power rule, we get: $\frac{dy}{dx} = -x^2 - 4x + 12$ Remember, the power rule says that the derivative of $x^n$ is $nx^{n-1}$ .
We bring the exponent down and reduce it by one.
We also know that the derivative of a constant (like $-3$ ) is just zero.

STEP 5

Now, we need to find the critical points by setting the derivative equal to zero and solving for $x$ .
This is where the slope is zero, which could indicate a peak or a valley.

STEP 6

So we have: $-x^2 - 4x + 12 = 0$ Multiplying by $-1$ to make it easier to factor, we get: $x^2 + 4x - 12 = 0$

STEP 7

This quadratic factors nicely as: $(x + 6)(x - 2) = 0$ So our critical points are $x = -6$ and $x = 2$ .

STEP 8

Now comes the fun part!
We need to test the intervals created by our critical points.
We have three intervals to consider: $(-\infty, -6)$ , $(-6, 2)$ , and $(2, \infty)$ .
We'll pick a test point in each interval and plug it into our derivative to see if it's positive or negative.

STEP 9

* Interval 1: $(-\infty, -6)$ .
Let's pick $x = -7$ .
Plugging it into the derivative, we get $-(-7)^2 - 4(-7) + 12 = -49 + 28 + 12 = -9$ .
This is negative, so the function is decreasing on this interval.

STEP 10

* Interval 2: $(-6, 2)$ .
Let's pick $x = 0$ .
Plugging it in, we get $-(0)^2 - 4(0) + 12 = 12$ .
This is positive, so the function is increasing on this interval.

STEP 11

* Interval 3: $(2, \infty)$ .
Let's pick $x = 3$ .
Plugging it in, we get $-(3)^2 - 4(3) + 12 = -9 - 12 + 12 = -9$ .
This is negative, so the function is decreasing on this interval.

STEP 12

(a) The function is increasing on $(-6, 2)$ . (b) The function is decreasing on $(-\infty, -6)$ and $(2, \infty)$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the derivative.
2. Find critical points.
3. Analyze intervals.

STEP 3

STEP 4

Using the power rule, we get: $\frac{dy}{dx} = -x^2 - 4x + 12$ Remember, the power rule says that the derivative of $x^n$ is $nx^{n-1}$ .
We bring the exponent down and reduce it by one.
We also know that the derivative of a constant (like $-3$ ) is just zero.

STEP 5

Now, we need to find the critical points by setting the derivative equal to zero and solving for $x$ .
This is where the slope is zero, which could indicate a peak or a valley.

STEP 6

So we have: $-x^2 - 4x + 12 = 0$ Multiplying by $-1$ to make it easier to factor, we get: $x^2 + 4x - 12 = 0$

STEP 7

This quadratic factors nicely as: $(x + 6)(x - 2) = 0$ So our critical points are $x = -6$ and $x = 2$ .

STEP 8

Now comes the fun part!
We need to test the intervals created by our critical points.
We have three intervals to consider: $(-\infty, -6)$ , $(-6, 2)$ , and $(2, \infty)$ .
We'll pick a test point in each interval and plug it into our derivative to see if it's positive or negative.

STEP 9

* Interval 1: $(-\infty, -6)$ .
Let's pick $x = -7$ .
Plugging it into the derivative, we get $-(-7)^2 - 4(-7) + 12 = -49 + 28 + 12 = -9$ .
This is negative, so the function is decreasing on this interval.

STEP 10

* Interval 2: $(-6, 2)$ .
Let's pick $x = 0$ .
Plugging it in, we get $-(0)^2 - 4(0) + 12 = 12$ .
This is positive, so the function is increasing on this interval.

STEP 11

* Interval 3: $(2, \infty)$ .
Let's pick $x = 3$ .
Plugging it in, we get $-(3)^2 - 4(3) + 12 = -9 - 12 + 12 = -9$ .
This is negative, so the function is decreasing on this interval.

STEP 12

(a) The function is increasing on $(-6, 2)$ . (b) The function is decreasing on $(-\infty, -6)$ and $(2, \infty)$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the derivative.2. Find critical points.3. Analyze intervals.

STEP 3

STEP 4

STEP 5

Now, we need to **find the critical points** by setting the derivative equal to zero and solving for xxx.This is where the slope is zero, which could indicate a peak or a valley.

STEP 6

So we have: −x2−4x+12=0 -x^2 - 4x + 12 = 0 −x2−4x+12=0 Multiplying by −1-1−1 to make it easier to factor, we get: x2+4x−12=0 x^2 + 4x - 12 = 0 x2+4x−12=0

STEP 7

This quadratic factors nicely as: (x+6)(x−2)=0 (x + 6)(x - 2) = 0 (x+6)(x−2)=0 So our **critical points** are x=−6x = -6x=−6 and x=2x = 2x=2.

STEP 8

STEP 9

STEP 10

* **Interval 2:** (−6,2)(-6, 2)(−6,2).Let's pick x=0x = 0x=0.Plugging it in, we get −(0)2−4(0)+12=12-(0)^2 - 4(0) + 12 = 12−(0)2−4(0)+12=12.This is **positive**, so the function is **increasing** on this interval.

STEP 11

* **Interval 3:** (2,∞)(2, \infty)(2,∞).Let's pick x=3x = 3x=3.Plugging it in, we get −(3)2−4(3)+12=−9−12+12=−9-(3)^2 - 4(3) + 12 = -9 - 12 + 12 = -9−(3)2−4(3)+12=−9−12+12=−9.This is **negative**, so the function is **decreasing** on this interval.

STEP 12

(a) The function is increasing on (−6,2)(-6, 2)(−6,2). (b) The function is decreasing on (−∞,−6)(-\infty, -6)(−∞,−6) and (2,∞)(2, \infty)(2,∞).

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1. Find the derivative.
2. Find critical points.
3. Analyze intervals.

Now, we need to find the critical points by setting the derivative equal to zero and solving for $x$ .
This is where the slope is zero, which could indicate a peak or a valley.

So we have: $-x^2 - 4x + 12 = 0$ Multiplying by $-1$ to make it easier to factor, we get: $x^2 + 4x - 12 = 0$

This quadratic factors nicely as: $(x + 6)(x - 2) = 0$ So our critical points are $x = -6$ and $x = 2$ .

* Interval 2: $(-6, 2)$ .
Let's pick $x = 0$ .
Plugging it in, we get $-(0)^2 - 4(0) + 12 = 12$ .
This is positive, so the function is increasing on this interval.

* Interval 3: $(2, \infty)$ .
Let's pick $x = 3$ .
Plugging it in, we get $-(3)^2 - 4(3) + 12 = -9 - 12 + 12 = -9$ .
This is negative, so the function is decreasing on this interval.

(a) The function is increasing on $(-6, 2)$ . (b) The function is decreasing on $(-\infty, -6)$ and $(2, \infty)$ .