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qijun's Question
Math
Posted 3 months ago
Use calculus methods to find the absolute maximum and absolute minimum values of the function
f(x) = \frac{x^3}{3} - 5 x^2 + 24 x
on the interval [-4,7).
(Hint: First solve the problem for [-4,7] using calculus. Then use a graph to figure out if removing the right endpoint makes a difference.) Enter -1000 for an absolute extremum that does not exist. 

Absolute maximum =  

Absolute minimum = 
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Answer from Sia
Posted 3 months ago
Solution by Steps
step 1
To find the critical points of the function f(x)=x335x2+24x f(x) = \frac{x^3}{3} - 5x^2 + 24x , we need to find the derivative and set it equal to zero
step 2
The derivative of f(x) f(x) is f(x)=x210x+24 f'(x) = x^2 - 10x + 24
step 3
Setting f(x)=0 f'(x) = 0 gives us the quadratic equation x210x+24=0 x^2 - 10x + 24 = 0
step 4
Factoring the quadratic equation, we get (x4)(x6)=0 (x - 4)(x - 6) = 0 , so the critical points are x=4 x = 4 and x=6 x = 6
step 5
Evaluate f(x) f(x) at the critical points and the endpoints of the interval [4,7][-4,7]
step 6
For x=4 x = -4 , f(4)=(4)335(4)2+24(4)=6438096=229763 f(-4) = \frac{(-4)^3}{3} - 5(-4)^2 + 24(-4) = \frac{-64}{3} - 80 - 96 = \frac{-22976}{3}
step 7
For x=4 x = 4 , f(4)=4335(4)2+24(4)=64380+96=643+16=1123 f(4) = \frac{4^3}{3} - 5(4)^2 + 24(4) = \frac{64}{3} - 80 + 96 = \frac{64}{3} + 16 = \frac{112}{3}
step 8
For x=6 x = 6 , f(6)=6335(6)2+24(6)=2163180+144=72+144=216 f(6) = \frac{6^3}{3} - 5(6)^2 + 24(6) = \frac{216}{3} - 180 + 144 = 72 + 144 = 216
step 9
For x=7 x = 7 , f(7)=7335(7)2+24(7)=3433245+168=343377=6863231=4553 f(7) = \frac{7^3}{3} - 5(7)^2 + 24(7) = \frac{343}{3} - 245 + 168 = \frac{343}{3} - 77 = \frac{686}{3} - 231 = \frac{455}{3}
step 10
Comparing the values, we find that the absolute maximum is f(6)=216 f(6) = 216 and the absolute minimum is f(4)=229763 f(-4) = \frac{-22976}{3}
step 11
Since the interval is [4,7)[-4,7), the value at x=7 x = 7 is not included. However, this does not affect the absolute maximum, which occurs at x=6 x = 6 . The absolute minimum is still at x=4 x = -4
Answer
Absolute maximum = 216
Absolute minimum = 229763\frac{-22976}{3}
Key Concept
Finding Absolute Extrema
Explanation
To find the absolute maximum and minimum values of a continuous function on a closed interval, we evaluate the function at critical points and endpoints, then compare these values. The highest value is the absolute maximum, and the lowest is the absolute minimum. If the interval is open at an endpoint, we exclude the value at that endpoint.

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