F(x)=x^3-3x^2+1 find the x value of all points where the function has a relative extrema. find the value(s) of any relative extrema

at x = 0 the relative extrema is 1

Step-by-step explanation:

The relative max and min occur when the derivative is zero.

Find the derivative of the function:

f(x) = x³ - 3x² + 1

f'(x) = 3x² - 6x + 0

Set the derivative equal to zero, factor, and solve for x:

3x² - 6x = 0

3x(x - 2) = 0

3x = 0    and    x - 2 = 0

 x = 0    and         x = 2

One of these is the max (extrema) and the other is the min (minima). Plug these x-values into the original equation to find the y-values.

f(0) = (0)³ - 3(0)² + 1

     =   0   -    0   + 1

     =   1

f(2) = (2)³ - 3(2)² + 1

     =  8    -  12    + 1

     =  -3

Since 1 is greater than -3, then 1 is the max and -3 is the min. (see attched graph).

there is no question content here, but you for the free points! : d