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Given the function y = x^4 - 8x^2 + 16. On which intervals is the function increasing?

A. Empty set
B. (-0,0)
C. (-[infinity], -2) and (0,2)
D. (-2, 0) and (2, 0)

1 Answer

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Final answer:

The function y = x⁴ - 8x² + 16 is increasing on the intervals (-2, 0) and (2, ∞), which is determined by setting the derivative to zero and testing the intervals between the critical points.

Step-by-step explanation:

To determine at which intervals the function y = x⁴ - 8x² + 16 is increasing, we need to find the derivative of the function and analyze the sign changes.

The derivative of the function is dy/dx = 4x³ - 16x. To find the critical points, we set the derivative to zero:

  • 4x³ - 16x = 0
  • x(4x² - 16) = 0
  • x(2x + 4)(2x - 4) = 0
  • x = 0, x = -2, x = 2

These are the critical points. To determine the intervals on which the function is increasing or decreasing, we create a sign chart and test values between the critical points:

  • For x < -2, choose x = -3: 4(-3)³ - 16(-3) = -108 + 48, which is negative. Therefore, the function is decreasing.
  • For -2 < x < 0, choose x = -1: 4(-1)³ - 16(-1) = -4 + 16, which is positive. Thus, the function is increasing.
  • For 0 < x < 2, choose x = 1: 4(1)³ - 16(1) = 4 - 16, which is negative. Hence, the function is decreasing.
  • For x > 2, choose x = 3: 4(3)³ - 16(3) = 108 - 48, which is positive. Therefore, the function is increasing.

From this analysis, the function is increasing on the intervals (-2, 0) and (2, ∞).

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