Find the critical points of f(x) = x^3 - 3x^2 + 4.

Find the critical points of f(x) = x^3 - 3x^2 + 4.

Correct Answer: x = 0 and x = 2

Step 1: Start with the function f(x) = x^3 - 3x^2 + 4.
Step 2: Find the derivative of the function, which is f'(x). The derivative of f(x) is f'(x) = 3x^2 - 6x.
Step 3: Set the derivative equal to zero to find critical points: 3x^2 - 6x = 0.
Step 4: Factor the equation: 3x(x - 2) = 0.
Step 5: Solve for x by setting each factor equal to zero: 3x = 0 gives x = 0, and x - 2 = 0 gives x = 2.
Step 6: The critical points are x = 0 and x = 2.
Step 7: To find the value of the function at a point, choose a point between the critical points, like x = 1.
Step 8: Evaluate f(1) = 1^3 - 3(1^2) + 4 = 1 - 3 + 4 = 2.

Finding Critical Points – This involves taking the derivative of a function, setting it to zero, and solving for x to find points where the function's slope is zero.
Evaluating Functions – This involves substituting values back into the original function to find corresponding y-values at the critical points.