Find the absolute minimum and absolute maximum values of f on the given interval.

Question

f(t) = t36 − t^2,    [−1, 6]

Answer 1

https://youtu.be/bu-FlcEup9E

Here is a link to a video explanation of how to solve this problem! I hope it helps! If there are any other video explanations you would like to see, please feel free to let me know in the comments. :) Thank you!

Answer 2

f(t) = t√(36 - t²)

f(t) is continuous on [-1,6], so f achieves absolute extreme values on the interval. These values occur at critical points or endpoints.

f'(t) = √(36 - t²) + (1/2)t(36 - t²)^-1/2(-2t) = √(36 - t²) - t² / √(36 - t²) = (36 - t² - t²) / √(36 - t²)

f'(t) = (36 - 2t²) / √(36 - t²) = 0 when t = √18 or -√18 (not in interval)

f(-1) = -√35 (absolute minimum value of f on [-1,6])

f(√18) = 18 (absolute maximum value of f on [-1,6])

f(6) = 0