Finding Absolute Extrema in Calculus

Finding Absolute Extrema

• Step 1: Find the critical points of the function within the given interval by differentiating the function and setting the derivative equal to zero.
• Step 2: Evaluate the function at the critical points and endpoints of the interval
• Step 3: The largest value obtained in Step 2 is the absolute maximum, and the smallest value is the absolute minimum.

Problem 1: Finding the Absolute Extrema of g(t) = 8t - t4

• The derivative of g(t) is g'(t) = 8 - 4t3.
• Setting g'(t) = 0, we get t = ³√2.
• Evaluating g(t) at the critical point (³√2) and endpoints (-2, 1), we get:
  • g(-2) = -16
  • g(³√2) ≈ 10.59
  • g(1) = 7
• Therefore, the absolute maximum of g(t) is approximately 10.59 at t = ³√2, and the absolute minimum is -16 at t = -2.

Problem 2: Finding the Absolute Extrema of g(x) = x²/³

• The derivative of g(x) is g'(x) = (2/3)x^(-1/3).
• While the derivative is undefined at x = 0, this point is within the interval.
• Evaluating g(x) at the critical point (0) and endpoints (-2, 3), we get:
  • g(-2) = 2^(2/3) ≈ 1.59
  • g(0) = 0
  • g(3) = 3^(2/3) ≈ 2.08
• Therefore, the absolute maximum of g(x) is approximately 2.08 at x = 3, and the absolute minimum is 0 at x = 0.