We are given the function:

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

We are asked to find the critical points and determine the behavior of the function at each of these points.

Step 1: Find the derivative of the function

To find the critical points, we first need to compute the first derivative of the function $f(x) = x^3 - 3x + 3$.

$f'(x) = 3x^2 - 3$

Step 2: Set the derivative equal to zero to find critical points

Critical points occur where the first derivative is zero or undefined. In this case, the derivative is a polynomial, so it is defined everywhere. We solve for $x$ by setting $f'(x) = 0$:

$3x^2 - 3 = 0$

$x^2 = 1$

$x = \pm 1$

Thus, the critical points are $x = 1$ and $x = -1$.

Step 3: Determine the behavior at each critical point (Second derivative test)

To determine whether the critical points are maxima, minima, or points of inflection, we use the second derivative test.

First, compute the second derivative:

$f''(x) = 6x$

• At $x = 1$: $f''(1) = 6(1) = 6 > 0$ Since the second derivative is positive, the function is concave up at $x = 1$, meaning this is a local minimum.

• At $x = -1$: $f''(-1) = 6(-1) = -6 < 0$ Since the second derivative is negative, the function is concave down at $x = -1$, meaning this is a local maximum.

Conclusion:

• There is a local minimum at $x = 1$.
• There is a local maximum at $x = -1$.

Would you like any further explanation or details on this solution?

Here are some related questions:

1. What is the significance of the second derivative in determining concavity?
2. How can you determine inflection points in a cubic function?
3. Can you apply the first derivative test to verify critical points instead of the second derivative?
4. How does the cubic nature of the function affect its graph?
5. What happens to the function as $x \to \infty$ or $x \to -\infty$?

Tip: Always confirm critical points using both the first and second derivative tests for a more complete understanding of the function’s behavior.