To find the maximum, minimum, and inflection points of the function $y = x^3 - 3x^2 + 4$, we will follow these steps:

1. First Derivative (To find critical points)

The first derivative $y'$ is the slope of the tangent to the curve at any point, and we use it to find critical points (where the slope is zero):

$y' = \frac{d}{dx} \left( x^3 - 3x^2 + 4 \right) = 3x^2 - 6x$

2. Set the First Derivative Equal to Zero (Critical points)

Set $y' = 0$ to find the critical points: $3x^2 - 6x = 0$ Factor out $3x$: $3x(x - 2) = 0$ So, the solutions are: $x = 0 \quad \text{or} \quad x = 2$ These are the critical points where we may have a local maximum or minimum.

3. Second Derivative (To classify critical points)

To classify these points as maxima, minima, or neither, we use the second derivative test. The second derivative $y''$ gives us the concavity of the function:

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

Now, evaluate the second derivative at the critical points:

• At $x = 0$: $y''(0) = 6(0) - 6 = -6$ Since $y''(0) < 0$, the function is concave down at $x = 0$, so there is a local maximum at $x = 0$.

• At $x = 2$: $y''(2) = 6(2) - 6 = 6$ Since $y''(2) > 0$, the function is concave up at $x = 2$, so there is a local minimum at $x = 2$.

4. Find the Inflection Point

The inflection point occurs where the second derivative changes sign (from positive to negative or vice versa). To find the inflection points, set $y'' = 0$:

$6x - 6 = 0$ Solving for $x$: $x = 1$ At $x = 1$, there is a possible inflection point.

5. Sketching the Graph

• Local Maximum: At $x = 0$, the value of $y$ is: $y(0) = 0^3 - 3(0)^2 + 4 = 4$ So there is a local maximum at $(0, 4)$.

• Local Minimum: At $x = 2$, the value of $y$ is: $y(2) = 2^3 - 3(2)^2 + 4 = 8 - 12 + 4 = 0$ So there is a local minimum at $(2, 0)$.

• Inflection Point: At $x = 1$, the value of $y$ is: $y(1) = 1^3 - 3(1)^2 + 4 = 1 - 3 + 4 = 2$ So there is an inflection point at $(1, 2)$.

6. Graph Overview

The graph of $y = x^3 - 3x^2 + 4$ has:

• A local maximum at $(0, 4)$,
• A local minimum at $(2, 0)$,
• An inflection point at $(1, 2)$.

Now, let's visualize the curve.

I'll sketch the graph for you.

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Do you have any specific details you'd like me to add, or shall I proceed with the sketch? Here are some related questions:

1. How does the first derivative test compare to the second derivative test in determining critical points?
2. Can a function have multiple inflection points?
3. What is the geometric interpretation of an inflection point?
4. How do you determine concavity of a function using the second derivative?
5. Can the second derivative test fail, and if so, what should be done?

Tip: Always check the sign of the second derivative around the inflection point to confirm a true change in concavity.