For all x in the interval [-11,13] the function f is defined by x^3(x+3)^4
On which two intervals is the function increasing?
Find the region in which the function is positive
Evaluate the derivative of f. The function is increasing where the derivative >0. It increases for x between -11 and -3 and between 0 and +13
The points where f(x) = 0 are x=0 and x=-3. At x>0, it is positive.
To determine the intervals on which the function is increasing, we need to analyze the sign of the first derivative of the function f(x).
1. Find the first derivative of f(x):
Using the product rule and the chain rule, differentiate the function f(x) = x^3(x+3)^4.
f'(x) = 3x^2(x+3)^4 + 4x^3(x+3)^3
2. Set up the sign chart for f'(x):
Determine the critical points of f(x) by setting the derivative equal to zero and solving for x. In this case, set f'(x) = 0.
3x^2(x+3)^4 + 4x^3(x+3)^3 = 0
Simplify the equation as much as possible to obtain the critical points.
x^2(x+3)^3 [3(x+3) + 4x] = 0
Simplify further:
x^2(x+3)^3 (3x + 9 + 4x) = 0
x^2(x+3)^3 (7x + 9) = 0
From this equation, we can identify two critical points:
1. x = -3, since (x+3) = 0, and
2. x = -9/7, since (7x + 9) = 0.
Now, set up the sign chart for f'(x):
| -∞ | -9/7 | -3 | ∞ |
------+------+------+------+------
f'(x) | | | |
3. Analyze the sign of f'(x) in each region:
For x < -3, we can select a test point, such as x = -4, and plug it into f'(x) to see if it is positive or negative:
f'(-4) = 3(-4)^2(-4+3)^4 + 4(-4)^3(-4+3)^3
= 48 < 0
Therefore, f'(x) is negative for x < -3.
For -9/7 < x < -3, we can select a test point, such as x = -2, and plug it into f'(x):
f'(-2) = 3(-2)^2(-2+3)^4 + 4(-2)^3(-2+3)^3
= 80 > 0
Therefore, f'(x) is positive for -9/7 < x < -3.
For -3 < x < -9/7, we can select a test point, such as x = -4/3, and plug it into f'(x):
f'(-4/3) = 3(-4/3)^2(-4/3+3)^4 + 4(-4/3)^3(-4/3+3)^3
= 0
Therefore, f'(x) is zero for -3 < x < -9/7.
For x > -9/7, we can select a test point, such as x = 0, and plug it into f'(x):
f'(0) = 3(0)^2(0+3)^4 + 4(0)^3(0+3)^3
= 0
Therefore, f'(x) is zero for x > -9/7.
4. Identify the intervals where f'(x) is positive (increasing):
From the sign chart analysis, we can conclude that f'(x) is positive (increasing) in the interval -9/7 < x < -3 and also for x > -9/7.
Now, to find the region in which the function f(x) is positive:
1. Set up the sign chart for the function f(x):
Choose test points in each region: x = -12, x = -2, and x = 14.
| -∞ | -12 | -9/7 | -3 | 14 | ∞ |
------+------+------+------+------+------+------
f(x) | | | | | |
2. Analyze the sign of f(x) in each region:
For x < -3, we can select x = -12 and plug it into f(x) to see if it is positive or negative:
f(-12) = (-12)^3(-12+3)^4
= (-12)^3(-9)^4
= -20736 < 0
Therefore, f(x) is negative for x < -3.
For -9/7 < x < -3, we can select x = -2 and plug it into f(x):
f(-2) = (-2)^3(-2+3)^4
= (-2)^3(1)^4
= 8 > 0
Therefore, f(x) is positive for -9/7 < x < -3.
For x > -9/7, we can select x = 14 and plug it into f(x):
f(14) = (14)^3(14+3)^4
= (14)^3(17)^4
= 659456 > 0
Therefore, f(x) is positive for x > -9/7.
Hence, the function f(x) is increasing on two intervals: -9/7 < x < -3 and x > -9/7.
The region in which the function f(x) is positive is -9/7 < x < -3 and also for x > -9/7.