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Determine the absolute extrema of [tex]f(x) = 6x - x^2 - 10[/tex] on the interval [-10, 7].

The absolute minimum of [tex]f[/tex] is:

A. 13 at [tex]x = 7[/tex]
B. -16 at [tex]x = -10[/tex]
C. -16 at [tex]x = 7[/tex]
D. 13 at [tex]x = -10[/tex]

Answer :

Final answer:

The absolute minimum of the function f(x) = 6x - x² - 10 on the interval [-10, 7] is -16 and occurs at x = -10. This is determined by finding the derivative, setting it to zero to find critical points, and evaluating the function at critical points and endpoints. Which corresponds to option b) -16 at ( x = -10 ).

Explanation:

To determine the absolute extrema of f(x) = 6x - x² - 10 on the interval [-10, 7], we will need to:

  1. Find the derivative of f(x) and solve for f'(x) = 0 to find critical points.
  2. Evaluate f(x) at the critical points and at the endpoints of the interval.
  3. Determine the highest and lowest values from step 2 to find the absolute maximum and minimum.

The derivative of f(x) is f'(x) = 6 - 2x. Setting the derivative equal to zero gives us the critical point x = 3. The value of f(3) is 6(3) - 3² - 10 = 8.

Now we evaluate f(x) at the endpoints: f(-10) = 6(-10) - (-10)² - 10 = -60 - 100 - 10 = -170, and f(7) = 6(7) - 7² - 10 = 42 - 49 - 10 = -17.

Comparing f(3), f(-10), and f(7), we see that the minimum value on the interval is -170 when x = -10, and the maximum value is 8 when x = 3. Therefore, the absolute minimum of f is -16 at x = -10.

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