Find the global maximum and minimum for the function on the closed interval.
\[
f(x)=x^{3}-3 x^{2}+5 \quad-1 \leq x \leq 3
\]
Answer
Compare the values to find the global maximum and minimum: \(\boxed{\text{Global Maximum: } x = 0, f(x) = 5}\) and \(\boxed{\text{Global Minimum: } x = 2, f(x) = 1}\)
Step 1 ：Find the critical points by taking the derivative of the function and setting it equal to zero: \(f'(x) = 3x^2 - 6x\)
Step 2 ：Solve for x: \(3x^2 - 6x = 0\) gives critical points \(x = 0\) and \(x = 2\)
Step 3 ：Evaluate the function at the critical points and endpoints: \(f(0) = 5\), \(f(2) = 1\), \(f(-1) = 1\), and \(f(3) = 5\)
Step 4 ：Compare the values to find the global maximum and minimum: \(\boxed{\text{Global Maximum: } x = 0, f(x) = 5}\) and \(\boxed{\text{Global Minimum: } x = 2, f(x) = 1}\)
