Determine the open intervals on which the function is increasing, decreasing, or constant.
f(x) = 3
f(x) = x^(2/3)
f(x) = -x^(3/4)
if f' > 0, f is increasing. So,
f(x)=3
f' = 0
f is constant everywhere
f(x) = x^(2/3)
f' = 2/3 x^(-1/3)
f' > 0, so f is increasing everywhere
f(x) = -x^(3/4)
f' = -3/4 x^(-1/4)
f' < 0 so, f is decreasing everywhere
rather poor examples, since there are no finite open intervals involved
To determine the open intervals on which a function is increasing, decreasing, or constant, we need to analyze the behavior of its derivative.
Let's find the derivatives of the given functions:
1. f(x) = 3
The derivative of a constant function is zero. So, f'(x) = 0. Therefore, this function is constant on its entire domain, and it never increases or decreases.
2. f(x) = x^(2/3)
To find the derivative, we can use the power rule. The power rule states that if we have a function in the form f(x) = x^n, its derivative is f'(x) = nx^(n-1).
Applying the power rule, we find:
f'(x) = (2/3)x^((2/3)-1) = (2/3)x^(-1/3) = 2/(3x^(1/3))
Now, let's analyze the sign of the derivative to determine the intervals of increase, decrease, or constancy.
Sign of f'(x):
- If f'(x) > 0, the function is increasing.
- If f'(x) < 0, the function is decreasing.
- If f'(x) = 0, the function is constant.
Since f'(x) = 2/(3x^(1/3)), we need to determine when f'(x) > 0, f'(x) < 0, and when f'(x) = 0.
Setting f'(x) > 0:
2/(3x^(1/3)) > 0
Since the denominator is always positive, we can multiply both sides by it without changing the inequality:
2 > 0
This inequality is always true. Therefore, f'(x) > 0 for all values of x. Hence, the function f(x) = x^(2/3) is increasing on its entire domain.
3. f(x) = -x^(3/4)
Following the same process as before, we find:
f'(x) = (-3/4)x^((3/4)-1) = (-3/4)x^(-1/4) = -3/(4x^(1/4))
Now, let's analyze the sign of the derivative to find the intervals of increase, decrease, or constancy.
Setting f'(x) > 0:
-3/(4x^(1/4)) > 0
Since the numerator (-3) is always negative, we need to consider the sign of the denominator x^(1/4):
- If x^(1/4) > 0 (x > 0), f'(x) < 0.
- If x^(1/4) < 0 (x < 0), f'(x) > 0.
Setting f'(x) = 0:
-3/(4x^(1/4)) = 0
There is no solution since the numerator is nonzero.
Overall, we have:
- f'(x) > 0 for x < 0, meaning that the function f(x) = -x^(3/4) is increasing on the interval (-β, 0).
- f'(x) < 0 for x > 0, meaning that the function f(x) = -x^(3/4) is decreasing on the interval (0, β).
Therefore, the function f(x) = -x^(3/4) is increasing on (-β, 0) and decreasing on (0, β).