Step 1 :Given the function \(f(x)=x^{3}-3 x^{2}+1\), we want to find any relative maxima and minima.
Step 2 :We start by taking the derivative of the function, \(f'(x) = 3x^{2} - 6x\).
Step 3 :We set the derivative equal to zero and solve for x to find the critical points, which gives us \(x = 0\) and \(x = 2\).
Step 4 :We then take the second derivative of the function, \(f''(x) = 6x - 6\).
Step 5 :We substitute the critical points into the second derivative. If the second derivative at a critical point is positive, the point is a relative minimum. If it's negative, the point is a relative maximum. If it's zero, the test is inconclusive.
Step 6 :Substituting \(x = 0\) into the second derivative gives a negative value, so \(x = 0\) is a relative maximum.
Step 7 :Substituting \(x = 2\) into the second derivative gives a positive value, so \(x = 2\) is a relative minimum.
Step 8 :\(\boxed{\text{Final Answer: The relative maximum of the function } f(x)=x^{3}-3 x^{2}+1 \text{ is at } x=0 \text{ and the relative minimum is at } x=2}\)