# Extreme point.Find the extreme point of f(x)=18x-3x^2.

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You should notice that the given function is a quadratic function, hence, the function reaches its extreme point at its vertex. Using the following formula, you may evaluate the vertex, such that:

`{(x = -b/(2a)),(y = -Delta/(4a)):}` => `{(x = -b/(2a)),(y = (4ac - b^2)/(4a)):}`

Identifying `a = -3, b = 18, c = 0` , yields:

`x = -18/(2*(-3)) => x = 3`

`y = (4*(-3)*0 - 18^2)/(4*(-3)) => y = 27`

**Hence, evaluating the vertex of the given function yields **`x = 3, y = 27.`

f(x)=18-3x^2.

First, we'll have to determine the critical values for f(x).

For this reason, we'll need to determine the first derivative:

f'(x) = 18-6x

The critical values of the function are the roots of the first derivative:

18-6x = 0

-6x = -18

x = 3

The given function has an extreme value for the critical value x = 3

Since the coefficient of x^2 is negative, then the function reaches the maximum values for x = 3

f(3) = 54 - 27 = 27

The maximum value of the function is f(3) = 27.