# Find the local maximum and minimum values of f(x) = x^3 -3x^2-9x where x lies in [-4, 4].

The minimum and maximum values of f(x) = x^3 - 3x^2 - 9x  have to be determined. At the points where the values are minimum and maximum f'(x) = 0. If f'(x) = 0 at x = c and f''(c) is negative there is a maximum value at c, else...

The minimum and maximum values of f(x) = x^3 - 3x^2 - 9x  have to be determined. At the points where the values are minimum and maximum f'(x) = 0. If f'(x) = 0 at x = c and f''(c) is negative there is a maximum value at c, else if f''(c) is positive there is a minimum value at x = c.

f(x) = x^3 - 3x^2 - 9x

f'(x) = 3x^2 - 6x - 9

Solving 3x^2 - 6x - 9 = 0

=> 3x^2 - 6x - 9 = 0

=> x^2 - 2x - 3 = 0

=> x^2 - 3x + x - 3 = 0

=> x(x - 3) + 1(x - 3) = 0

=> (x + 1)(x - 3) = 0

=> x = -1 and x = 3

Both x = -1 and x= 3 lie in [-4, 4].

The maximum value is f(-1) = 5 and the minimum value is f(-3) = -27.

It is not -76 as can be seen from the graph below:

The correct minimum value of the function for x in [-4, 4] is -27 and the maximum value is 5.

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You need to calculate the derivative of the function to find the maximum and minimum values of the function.

f'(x) = `3x^2 - 6x - 9`

You need to find the zeroes of derivative:

f'(x) = 0 => `3x^2 - 6x - 9`  = 0 =>`x^2 - 2x - 3 ` = 0

`x^2 - 2x - 2 -1 = 0`  => (x^2 - 1) - 2(x+1) = 0

(x-1)(x+1) - 2(x+1) = 0

Factoring out (x+1) yields:

(x+1)(x - 3) = 0 => x + 1 = 0 => x = -1

x - 3 = 0 => x = 3

The values of derivative of the function are negative between -1 and 3, hence the function decreases between -1 and 3.

The function attends its maximum value at x = -1 => f(-1) = -1 - 3 + 9 = 5.

The function attends its minimum value at x = 3 => f(3) = 27 - 27 - 27 = -27.

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