Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = 3 (sqaure root sign) t (20 − t), [0,

Question

3 years ago

9

20]

1 answer:

mafiozo [28] · Answer 1 · 2022-09-02T14:38:31+03:00

9514 1404 393

Answer:

Step-by-step explanation:

The minimum will be found at the ends of the interval, where f(t) = 0.

The maximum is found in the middle of the interval, where f'(t) = 0.

$f(t)=\sqrt[3]{t}(20-t)\\\\f'(t)=\dfrac{20-t}{3\sqrt[3]{t^2}}-\sqrt[3]{t}=\sqrt[3]{t}\left(\dfrac{4(5-t)}{3t}\right)$

This derivative is zero when the numerator is zero, at t=5. The function is a maximum at that point. The value there is ...

f(5) = (∛5)(20-5) = 15∛5

The absolute maximum on the interval is 15∛5 at t=5.

Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = 3 (*sqaure root sign*) t (20 − t), [0,