Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function y = x^3 - 3x + 3.

Steps to Solve

1. Find the first derivative

To locate local and absolute extreme points, we start by finding the first derivative of the function ( y = x^3 - 3x + 3 ):

[ y' = \frac{d}{dx}(x^3 - 3x + 3) = 3x^2 - 3 ]

2. Set the first derivative to zero

Next, we set the first derivative equal to zero to find the critical points:

[ 3x^2 - 3 = 0 ]

Dividing by 3:

[ x^2 - 1 = 0 ]

Factoring the equation:

[ (x - 1)(x + 1) = 0 ]

Thus, the critical points are:

[ x = 1 \quad \text{and} \quad x = -1 ]

3. Determine the second derivative

Now, for classification of the critical points, we'll find the second derivative:

[ y'' = \frac{d^2}{dx^2}(x^3 - 3x + 3) = 6x ]

4. Evaluate the second derivative at critical points

We evaluate ( y'' ) at the critical points to determine their nature:

• For ( x = 1 ):

[ y''(1) = 6(1) = 6 \quad (\text{Local minimum}) ]

• For ( x = -1 ):

[ y''(-1) = 6(-1) = -6 \quad (\text{Local maximum}) ]

5. Find the coordinates of the critical points

Now we find the ( y )-coordinates at the critical points:

• For ( x = 1 ):

[ y(1) = (1)^3 - 3(1) + 3 = 1 ]

• For ( x = -1 ):

[ y(-1) = (-1)^3 - 3(-1) + 3 = 5 ]

Thus, the local maximum is at ( (-1, 5) ) and the local minimum is at ( (1, 1) ).

6. Find inflection points

To find inflection points, we set the second derivative equal to zero:

[ 6x = 0 \Rightarrow x = 0 ]

Evaluating ( y ) at ( x = 0 ):

[ y(0) = 0^3 - 3(0) + 3 = 3 ]

So, the inflection point is at ( (0, 3) ).

7. Summarize points

• Local maximum: ( (-1, 5) )
• Local minimum: ( (1, 1) )
• Inflection point: ( (0, 3) )

8. Graphing the function

Using graphing tools, we can visualize the function ( y = x^3 - 3x + 3 ) along with the identified points.