For the function f(x) = x^4 - 8x^2 + 16, find the intervals where the function is increasing.

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Q: For the function f(x) = x^4 - 8x^2 + 16, find the intervals where the function is increasing.

Solution: f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x(x^2 - 4) = 0, so x = -2, 0, 2. Test intervals: f' is positive in (-2, ∞).

Steps: 12

Step 1: Identify the function we are working with, which is f(x) = x^4 - 8x^2 + 16.
Step 2: Find the derivative of the function, f'(x), to determine where the function is increasing or decreasing. The derivative is f'(x) = 4x^3 - 16x.
Step 3: Set the derivative equal to zero to find critical points: 4x^3 - 16x = 0.
Step 4: Factor the equation: 4x(x^2 - 4) = 0.
Step 5: Solve for x: This gives us x = 0, x = -2, and x = 2 as critical points.
Step 6: Determine the intervals to test: The critical points divide the number line into intervals: (-∞, -2), (-2, 0), (0, 2), and (2, ∞).
Step 7: Choose a test point from each interval and plug it into the derivative f'(x) to see if it is positive (increasing) or negative (decreasing).
Step 8: For the interval (-∞, -2), test x = -3: f'(-3) = 4(-3)^3 - 16(-3) = -108 + 48 = -60 (negative).
Step 9: For the interval (-2, 0), test x = -1: f'(-1) = 4(-1)^3 - 16(-1) = -4 + 16 = 12 (positive).
Step 10: For the interval (0, 2), test x = 1: f'(1) = 4(1)^3 - 16(1) = 4 - 16 = -12 (negative).
Step 11: For the interval (2, ∞), test x = 3: f'(3) = 4(3)^3 - 16(3) = 108 - 48 = 60 (positive).
Step 12: Summarize the results: The function is increasing in the intervals (-2, 0) and (2, ∞).