Identify the coordinates of any local maximum and minimum points for the function y = x^3 - 3x + 3 and select the correct answers from the options given.

Steps to Solve

1. Find the derivative of the function To identify local maximum and minimum points, we first need the derivative of the function. Given the function

$$ y = x^3 - 3x + 3 $$

the derivative is:

$$ y' = 3x^2 - 3 $$

2. Set the derivative to zero Next, we set the derivative equal to zero to find critical points:

$$ 3x^2 - 3 = 0 $$

This simplifies to:

$$ x^2 = 1 $$

3. Solve for critical points By taking the square root, we find the critical points:

$$ x = 1 \quad \text{and} \quad x = -1 $$

4. Determine the nature of the critical points To classify the critical points, we can use the second derivative test. We calculate the second derivative:

$$ y'' = 6x $$

Now we evaluate the second derivative at the critical points:

• For $x = 1$: $$ y''(1) = 6(1) = 6 \quad (\text{positive, indicates a local minimum}) $$

• For $x = -1$: $$ y''(-1) = 6(-1) = -6 \quad (\text{negative, indicates a local maximum}) $$

5. Find the function values at critical points Now, we evaluate the original function at these critical points to find their coordinates:

• For $x = -1$: $$ y(-1) = (-1)^3 - 3(-1) + 3 = -1 + 3 + 3 = 5 $$ Thus, the local maximum point is at $(-1, 5)$.

• For $x = 1$: $$ y(1) = (1)^3 - 3(1) + 3 = 1 - 3 + 3 = 1 $$ Thus, the local minimum point is at $(1, 1)$.