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.
![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)](https://tex.z-dn.net/?f=f%28t%29%3D%5Csqrt%5B3%5D%7Bt%7D%2820-t%29%5C%5C%5C%5Cf%27%28t%29%3D%5Cdfrac%7B20-t%7D%7B3%5Csqrt%5B3%5D%7Bt%5E2%7D%7D-%5Csqrt%5B3%5D%7Bt%7D%3D%5Csqrt%5B3%5D%7Bt%7D%5Cleft%28%5Cdfrac%7B4%285-t%29%7D%7B3t%7D%5Cright%29)
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.
2:12
it's Clara because if you simplify it's 6:1 flip sides
<span>In 3 years, you will have $8,103.38
Formula:
</span>
<span>A = P (1 + r/n)<span> ^(nt)</span></span>
Where:
A = the future value of the investment/loan, including interest
P = the principal investment amount (the initial deposit or loan amount)
r = the annual interest rate (decimal)
n = the number of times that interest is compounded per year
t = the number of years the money is invested or borrowed for
