Final answer:
To find the intervals on which the function f(x) = x + 4/x^2 is increasing or decreasing, we calculate the first derivative and analyze its sign. We find that f(x) is decreasing on the interval (0, 2) and increasing on the interval (2, ∞). The function has no local maximum or minimum values.
Step-by-step explanation:
To determine the intervals on which the function f(x) = x + 4/x^2 is increasing or decreasing, we need to find the first derivative of the function and analyze its sign. Let's find the derivative:
f'(x) = 1 - 8/x^3
To find the critical points, we set f'(x) = 0 and solve for x:
1 - 8/x^3 = 0
8/x^3 = 1
x^3 = 8
x = 2
To determine the intervals, we can choose a test point in each interval and evaluate f'(x) for that point. For example, if we choose x = 1, then f'(1) = 1 - 8/1^3 = 1 - 8 = -7, which is negative. This means that f(x) is decreasing on the interval (0, 2).
Similarly, if we choose x = 3, then f'(3) = 1 - 8/3^3 = 1 - 8/27 = 19/27, which is positive. This means that f(x) is increasing on the interval (2, ∞).
To find the local maximum and minimum values, we need to find the second derivative of the function. Let's find it:
f''(x) = 24/x^4
Since the second derivative is always positive, this means that f(x) has no local maximum or minimum values.