find the relative extrema of the function f(x)= (x^3)-(6x^2)+(9x)+4. ?

To find the relative extrema of the function f(x) = x^3 - 6x^2 + 9x + 4, we first need to find the critical points by taking the derivative of the function and setting it equal to zero.

f'(x) = 3x^2 - 12x + 9

Setting f'(x) = 0, we get:

3x^2 - 12x + 9 = 0
x^2 - 4x + 3 = 0
(x - 3)(x - 1) = 0

So, x = 3 and x = 1 are the critical points.

To determine the nature of the critical points, we can use the second derivative test. Taking the second derivative of the function:

f''(x) = 6x - 12

Plugging in the critical points x = 3 and x = 1:

f''(3) = 6(3) - 12 = 6 > 0
f''(1) = 6(1) - 12 = -6 < 0

Since f''(3) > 0, x = 3 is a local minimum. And since f''(1) < 0, x = 1 is a local maximum.

Therefore, the relative extrema of the function f(x) = x^3 - 6x^2 + 9x + 4 are:

Local minimum at (3, 4)
Local maximum at (1, 12)