Answer :
- The maximum value is -1060, which occurs at x = -4.
- The minimum value is -1645, which occurs at x = 5.
To find the absolute extrema of the function f(x) = x - 66x^2 on the interval [-4, 5], we need to evaluate the function at both the endpoints and the critical points within the interval.
1. Evaluating at the endpoints:
- When x = -4, f(-4) = -4 - 66(-4)^2 = -4 - 66(16) = -4 - 1056 = -1060.
- When x = 5, f(5) = 5 - 66(5)^2 = 5 - 66(25) = 5 - 1650 = -1645.
2. Finding the critical points within the interval:
To find the critical points, we need to determine where the derivative of the function is equal to zero or undefined.
- First, let's find the derivative of f(x): f'(x) = 1 - 132x.
- Setting the derivative equal to zero: 1 - 132x = 0.
- Solving for x: 132x = 1, x = 1/132.
Since the interval [-4, 5] includes the critical point x = 1/132, we need to evaluate the function at this point as well.
- When x = 1/132, f(1/132) = (1/132) - 66(1/132)^2 = 1/132 - 66/17424 = 1/132 - 1/264 = 1/132 - 2/264 = 1/132 - 1/132 = 0.
Now, let's summarize the findings:
- At the endpoint x = -4, f(-4) = -1060.
- At the endpoint x = 5, f(5) = -1645.
- At the critical point x = 1/132, f(1/132) = 0.
Based on these evaluations, we can determine the absolute extrema:
- The maximum value is -1060, which occurs at x = -4.
- The minimum value is -1645, which occurs at x = 5.
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