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.
Answer:
Answer: 3x + 3 = 2x + 1
Answer: 3x + 3 = 2x + 1 <=>3x - 2x = 1 - 3
Answer: 3x + 3 = 2x + 1 <=>3x - 2x = 1 - 3 <=> x = -2
<span>Changing the grouping of the addends should not change the sum, according to the associative property of addition. You might group them differently with (50 + 3) + 47, so that you have 50 + (3 + 47). You might not regroup them with (16 + 4) + 5 rather than 16 + (4 + 5).</span>
The answer should be 1003.53