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Find the intervals on which (x) is increasing or decreasing, and find the local maximum and minimum values of (x) for: f(x)= x+ 4/ˣ²

User Sisley
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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.

User Galinette
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7.4k points
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