f is increasing where f' > 0
since the even powers are always positive, f'>0 where
x-3 > 0
and
x-4 > 0
so, that would be
(3,4)U(4,∞)
f '(x) = (x + 2)^4(x − 3)^7(x − 4)^6.
On what interval is f increasing? (Enter your answer in interval notation.)
since the even powers are always positive, f'>0 where
x-3 > 0
and
x-4 > 0
so, that would be
(3,4)U(4,∞)
Let's break it down: We have (x + 2)^4, (x - 3)^7, and (x - 4)^6. For f '(x) to be positive, each of these factors must be either positive or negative with an odd exponent.
(x + 2)^4 is positive for all x since it's raised to an even power.
(x - 3)^7 is positive when x > 3 because it's raised to an odd power and becomes negative when x < 3.
(x - 4)^6 is positive for all x since it's raised to an even power.
So combining all the information, we find that f '(x) is positive when x > 3 since (x - 3)^7 is positive in that interval.
Therefore, the function f is increasing on the interval (3, ∞).
Given that f'(x) = (x + 2)^4(x − 3)^7(x − 4)^6, the function f is increasing wherever its derivative is positive.
We can determine the sign of f'(x) by finding the x-values that make each factor positive, negative, or zero.
1. (x + 2)^4: This factor is positive for all x values, as it is a perfect fourth power.
2. (x − 3)^7: This factor is positive for x > 3, negative for x < 3, and zero at x = 3.
3. (x − 4)^6: This factor is positive for x > 4, negative for x < 4, and zero at x = 4.
Now let's analyze the overall sign of f'(x) by considering the signs of each factor:
Since (x + 2)^4 is always positive and does not affect the overall sign, we can focus on the other two factors.
- For (x − 3)^7, it changes sign at x = 3.
- For (x − 4)^6, it changes sign at x = 4.
By considering the sign changes at x = 3 and x = 4, we can determine the intervals where f'(x) is positive, and consequently, where f(x) is increasing.
From left to right, the intervals are:
1. (-∞, 3): f'(x) is negative because both factors are negative.
2. (3, 4): f'(x) is positive because (x − 3)^7 is positive, and (x − 4)^6 is negative.
3. (4, ∞): f'(x) is positive because both factors are positive.
Therefore, the interval on which f is increasing is (3, ∞) in interval notation.
The given derivative function is f'(x) = (x + 2)^4(x − 3)^7(x − 4)^6.
To find the critical points, we need to set f'(x) equal to zero and solve for x. However, in this case, we have a product of factors, and for the derivative to be zero, at least one of these factors must be equal to zero.
Setting each factor equal to zero, we find three possible critical points:
1) Factor 1: (x + 2)^4 = 0 -> x + 2 = 0 -> x = -2
2) Factor 2: (x − 3)^7 = 0 -> x - 3 = 0 -> x = 3
3) Factor 3: (x − 4)^6 = 0 -> x - 4 = 0 -> x = 4
Now, we need to test the intervals created by these critical points to determine if they are increasing or decreasing. We can use test points within each interval to do this.
For the interval (-∞, -2):
Choose a test point x = -3, substitute it into f'(x):
f'(-3) = (-3 + 2)^4(-3 − 3)^7(-3 − 4)^6 = (-1)^4(-6)^7(-7)^6
Since the value of f'(-3) is negative, the function is decreasing in this interval.
For the interval (-2, 3):
Choose a test point x = 0, substitute it into f'(x):
f'(0) = (0 + 2)^4(0 − 3)^7(0 − 4)^6 = (2)^4(-3)^7(-4)^6
Since the value of f'(0) is positive, the function is increasing in this interval.
For the interval (3, 4):
Choose a test point x = 3.5, substitute it into f'(x):
f'(3.5) = (3.5 + 2)^4(3.5 − 3)^7(3.5 − 4)^6 = (5.5)^4(0.5)^7(-0.5)^6
Since the value of f'(3.5) is negative, the function is decreasing in this interval.
For the interval (4, ∞):
Choose a test point x = 5, substitute it into f'(x):
f'(5) = (5 + 2)^4(5 − 3)^7(5 − 4)^6 = (7)^4(2)^7(1)^6
Since the value of f'(5) is positive, the function is increasing in this interval.
Therefore, the function f(x) is increasing on the interval (-2, 3) ∪ (4, ∞).