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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 $f(x) = \frac{x^3}{3} - 5x^2 + 24x$, we need to find the derivative and set it equal to zero

step 2

The derivative of $f(x)$ is $f'(x) = x^2 - 10x + 24$

step 3

Setting $f'(x) = 0$ gives us the quadratic equation $x^2 - 10x + 24 = 0$

step 4

Factoring the quadratic equation, we get $(x - 4)(x - 6) = 0$, so the critical points are $x = 4$ and $x = 6$

step 5

Evaluate $f(x)$ at the critical points and the endpoints of the interval $[-4,7]$

step 6

For $x = -4$, $f(-4) = \frac{(-4)^3}{3} - 5(-4)^2 + 24(-4) = \frac{-64}{3} - 80 - 96 = \frac{-22976}{3}$

step 7

For $x = 4$, $f(4) = \frac{4^3}{3} - 5(4)^2 + 24(4) = \frac{64}{3} - 80 + 96 = \frac{64}{3} + 16 = \frac{112}{3}$

step 8

For $x = 6$, $f(6) = \frac{6^3}{3} - 5(6)^2 + 24(6) = \frac{216}{3} - 180 + 144 = 72 + 144 = 216$

step 9

For $x = 7$, $f(7) = \frac{7^3}{3} - 5(7)^2 + 24(7) = \frac{343}{3} - 245 + 168 = \frac{343}{3} - 77 = \frac{686}{3} - 231 = \frac{455}{3}$

step 10

Comparing the values, we find that the absolute maximum is $f(6) = 216$ and the absolute minimum is $f(-4) = \frac{-22976}{3}$

step 11

Since the interval is $[-4,7)$, the value at $x = 7$ is not included. However, this does not affect the absolute maximum, which occurs at $x = 6$. The absolute minimum is still at $x = -4$

Answer

Absolute maximum = 216

Absolute minimum = $\frac{-22976}{3}$

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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