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

Tushar Chandra | Certified Educator

calendarEducator since 2010

starTop subjects are Math, Science, and Business

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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## Related Questions

Luca B. | Certified Educator

calendarEducator since 2011

starTop subjects are Math, Science, and Business

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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