Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = 3 (*sqaure root sign*) t (20 − t), [0,
1 answer:
9514 1404 393
Answer:
- maximum: 15∛5 ≈ 25.6496392002
- minimum: 0
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.
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.
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Answer:
- quotient: 10x +16
- remainder: 28x^2 +10x +22
Step-by-step explanation:
The attachment shows the steps.
quotient: 10x +16
remainder: 28x^2 +10x +22
_____
Polynomial long division is easier than numerical long division because you only need to examine the first terms to determine the quotient term.
The first quotient term is 10x^4/x^3 = 10x.
The second quotient term is 16x^3/x^3 = 16.
When the leading dividend term is lower degree than the divisor, that dividend is the remainder.
