Suppose that f is a differentiable function with derivative

Question

Suppose that f is a differentiable function with derivative

𝑓'(𝑥) = (𝑥 − 3)(𝑥 + 1)(𝑥 + 5). Determine the intervals of x for which the function of f is increasing and decreasing.

Answer 1

Well, well, well, looks like we have a differentiable function with the derivative (𝑥 − 3)(𝑥 + 1)(𝑥 + 5). To find the intervals of x where f is increasing and decreasing, we need to look at the sign of the derivative.

Let's break it down. The derivative (𝑥 − 3)(𝑥 + 1)(𝑥 + 5) is a product of three factors. We need to determine the intervals of x where each factor is positive, negative, or zero.

First, let's find when (𝑥 − 3) = 0. Solving this equation, we get 𝑥 = 3. So, (𝑥 − 3) is zero when x = 3.

Next, let's find when (𝑥 + 1) = 0. Solving this equation, we get 𝑥 = -1. So, (𝑥 + 1) is zero when x = -1.

Finally, let's find when (𝑥 + 5) = 0. Solving this equation, we get 𝑥 = -5. So, (𝑥 + 5) is zero when x = -5.

Now, let's consider each factor:

When 𝑥 < -5, all three factors are negative. So, (𝑥 − 3)(𝑥 + 1)(𝑥 + 5) is negative.

When -5 < 𝑥 < -1, (𝑥 − 3) is negative, (𝑥 + 1) is positive, and (𝑥 + 5) is negative. Therefore, (𝑥 − 3)(𝑥 + 1)(𝑥 + 5) is positive.

When -1 < 𝑥 < 3, (𝑥 − 3) is negative, (𝑥 + 1) is positive, and (𝑥 + 5) is positive. Hence, (𝑥 − 3)(𝑥 + 1)(𝑥 + 5) is negative.

Lastly, when 𝑥 > 3, all three factors are positive. So, (𝑥 − 3)(𝑥 + 1)(𝑥 + 5) is positive.

Now, remembering that f'(𝑥) is the derivative of f, we can determine the intervals of x for which the function f is increasing and decreasing.

When (𝑥 − 3)(𝑥 + 1)(𝑥 + 5) is positive, f'(𝑥) > 0, which means f is increasing.

When (𝑥 − 3)(𝑥 + 1)(𝑥 + 5) is negative, f'(𝑥) < 0, which means f is decreasing.

So, putting it all together, we have:

- f is increasing when x < -5 or -1 < x < 3.
- f is decreasing when -5 < x < -1 or x > 3.

I hope this answer tickled your mathematical funny bone!

Answer 2

To determine the intervals where the function f is increasing and decreasing, we need to analyze the sign of the derivative.

Given:

f'(x) = (x - 3)(x + 1)(x + 5)

We need to find the values of x for which the derivative is positive (increasing) and negative (decreasing).

Let's break down the problem step by step:

1. Find the critical points:
The critical points are the values of x for which f'(x) = 0 or f'(x) is undefined. In this case, f'(x) is defined for all real numbers, so we need to find where f'(x) = 0.

Setting f'(x) = 0:
(x - 3)(x + 1)(x + 5) = 0

This equation gives us three critical points: x = 3, x = -1, and x = -5.

2. Determine the intervals:
Now, we will divide the number line into four intervals using the critical points:
(-∞, -5), (-5, -1), (-1, 3), and (3, +∞).

3. Analyze the sign of f'(x) in each interval:
For each interval, we choose a test point to determine the sign of f'(x). A convenient choice is the midpoint of each interval.

- For x < -5, let's choose x = -6.
Plugging x = -6 into f'(x) = (x - 3)(x + 1)(x + 5), we get:
f'(-6) = (-6 - 3)(-6 + 1)(-6 + 5) = (-9)(-5)(-1) = 45 > 0

- For -5 < x < -1, let's choose x = -2.
Plugging x = -2 into f'(x) = (x - 3)(x + 1)(x + 5), we get:
f'(-2) = (-2 - 3)(-2 + 1)(-2 + 5) = (-5)(-1)(3) = 15 < 0

- For -1 < x < 3, let's choose x = 0.
Plugging x = 0 into f'(x) = (x - 3)(x + 1)(x + 5), we get:
f'(0) = (0 - 3)(0 + 1)(0 + 5) = (-3)(1)(5) = -15 < 0

- For x > 3, let's choose x = 4.
Plugging x = 4 into f'(x) = (x - 3)(x + 1)(x + 5), we get:
f'(4) = (4 - 3)(4 + 1)(4 + 5) = (1)(5)(9) = 45 > 0

4. Analyze the results:
Based on the signs of f'(x) in each interval, we can determine when the function f is increasing and decreasing.

- The function f is increasing on the intervals:
(-∞, -5) and (3, +∞)

- The function f is decreasing on the interval:
(-5, -1)

Therefore, the intervals where the function f is increasing are (-∞, -5) and (3, +∞), and the interval where f is decreasing is (-5, -1).

Answer 3

To determine the intervals on which the function f is increasing or decreasing, we need to analyze the sign of the derivative 𝑓'(𝑥).

1. Find the critical points:
To find the critical points, we need to solve the equation 𝑓'(𝑥) = 0.
Set (𝑥 − 3)(𝑥 + 1)(𝑥 + 5) = 0 and solve for 𝑥:
(𝑥 − 3)(𝑥 + 1)(𝑥 + 5) = 0
Setting each factor to zero, we get:
𝑥 − 3 = 0 => 𝑥 = 3
𝑥 + 1 = 0 => 𝑥 = -1
𝑥 + 5 = 0 => 𝑥 = -5

So, the critical points are 𝑥 = 3, 𝑥 = -1, and 𝑥 = -5.

2. Test intervals using test points:
Now, let's choose some test points in each interval determined by the critical points to determine the sign of the derivative 𝑓'(𝑥) in those intervals.

- For 𝑥 < -5, choose a test point 𝑥 = -10:
Plug 𝑥 = -10 into 𝑓'(𝑥):
𝑓'(-10) = (-10 − 3)(-10 + 1)(-10 + 5) = (-13)(-9)(-5) = -585

Since 𝑓'(-10) = -585 is negative, 𝑓(x) is decreasing in the interval 𝑥 < -5.

- For -5 < 𝑥 < -1, choose a test point 𝑥 = -3:
Plug 𝑥 = -3 into 𝑓'(𝑥):
𝑓'(-3) = (-3 − 3)(-3 + 1)(-3 + 5) = (-6)(-2)(2) = 24

Since 𝑓'(-3) = 24 is positive, 𝑓(x) is increasing in the interval -5 < 𝑥 < -1.

- For -1 < 𝑥 < 3, choose a test point 𝑥 = 0:
Plug 𝑥 = 0 into 𝑓'(𝑥):
𝑓'(0) = (0 − 3)(0 + 1)(0 + 5) = (-3)(1)(5) = -15

Since 𝑓'(0) = -15 is negative, 𝑓(x) is decreasing in the interval -1 < 𝑥 < 3.

- For 𝑥 > 3, choose a test point 𝑥 = 5:
Plug 𝑥 = 5 into 𝑓'(𝑥):
𝑓'(5) = (5 − 3)(5 + 1)(5 + 5) = (2)(6)(10) = 120

Since 𝑓'(5) = 120 is positive, 𝑓(x) is increasing in the interval 𝑥 > 3.

So, the summary of our findings is:
- 𝑓(x) is decreasing in the interval 𝑥 < -5.
- 𝑓(x) is increasing in the interval -5 < 𝑥 < -1.
- 𝑓(x) is decreasing in the interval -1 < 𝑥 < 3.
- 𝑓(x) is increasing in the interval 𝑥 > 3.

Answer 4

since f'(x) =0 at x = -5, -1, 3 f' changes sign there.

Since f'(-6) < 0,
f is increasing (f' > 0) in (-5,-1) and (3,∞)
f is decreasing on (-∞,-5) and (-1,3)

This is just Algebra II. Recall what you know about the properties of polynomials.