f(x)=3x4−16x3+6x2+72x−7
  

has critical points at some of the following x
-values. In each case identify whether the point is critical, and if so find out if it is a maximum, minimum or neither maximum nor minimum.

To identify the critical points of the function \( f(x) = 3x^4 - 16x^3 + 6x^2 + 72x - 7 \), we need to follow these steps:

1. **Find the first derivative \( f'(x) \)**.
2. **Set \( f'(x) = 0 \) and solve for \( x \) to find the critical points**.
3. **Determine the nature of each critical point** (whether it is a maximum, minimum, or neither) by analyzing the second derivative \( f''(x) \).

### Step 1: Find the First Derivative

Let's find the first derivative \( f'(x) \):

\[
f'(x) = \frac{d}{dx}(3x^4 - 16x^3 + 6x^2 + 72x - 7)
\]

Calculating each term:

\[
f'(x) = 12x^3 - 48x^2 + 12x + 72
\]

### Step 2: Set \( f'(x) = 0 \)

Set the first derivative equal to zero and solve for \( x \):

\[
12x^3 - 48x^2 + 12x + 72 = 0
\]

To simplify, divide the entire equation by 12:

\[
x^3 - 4x^2 + x + 6 = 0
\]

This is a cubic equation, which may require factorization or numerical methods to solve.

### Step 3: Solve the Cubic Equation

Let's solve \( x^3 - 4x^2 + x + 6 = 0 \) by trying to factor or use numerical methods.

To factor, we can check possible rational roots, such as \( x = 1, -1, 2, -2, 3, -3, 6, -6 \).

By substituting, we find that \( x = -1 \) is a root:

\[
x^3 - 4x^2 + x + 6 = (x + 1)(x^2 - 5x + 6)
\]

Now, factor the quadratic \( x^2 - 5x + 6 = 0 \):

\[
x^2 - 5x + 6 = (x - 2)(x - 3)
\]

So the complete factorization is:

\[
x^3 - 4x^2 + x + 6 = (x + 1)(x - 2)(x - 3)
\]

The solutions are \( x = -1 \), \( x = 2 \), and \( x = 3 \).

### Step 4: Determine the Nature of Each Critical Point

To determine whether each critical point is a maximum, minimum, or neither, we analyze the second derivative \( f''(x) \).

First, let's find the second derivative:

\[
f''(x) = \frac{d}{dx}(12x^3 - 48x^2 + 12x + 72)
\]

Calculating each term:

\[
f''(x) = 36x^2 - 96x + 12
\]

Now, substitute each critical point into \( f''(x) \):

- For \( x = -1 \):

\[
f''(-1) = 36(-1)^2 - 96(-1) + 12 = 36 + 96 + 12 = 144 > 0
\]

So, \( x = -1 \) is a **local minimum**.

- For \( x = 2 \):

\[
f''(2) = 36(2)^2 - 96(2) + 12 = 144 - 192 + 12 = -36 < 0
\]

So, \( x = 2 \) is a **local maximum**.

- For \( x = 3 \):

\[
f''(3) = 36(3)^2 - 96(3) + 12 = 324 - 288 + 12 = 48 > 0
\]

So, \( x = 3 \) is a **local minimum**.

### Final Conclusion:

The function \( f(x) = 3x^4 - 16x^3 + 6x^2 + 72x - 7 \) has critical points at:

- \( x = -1 \) (local minimum)
- \( x = 2 \) (local maximum)
- \( x = 3 \) (local minimum)

Would you like more details or have any questions about this solution? Here are some related questions to consider:

1. How do you find critical points for other types of functions?
2. What are the steps to solve higher-degree polynomials for critical points?
3. Can you always use the second derivative test to classify critical points?
4. How does the behavior of a cubic function differ from that of a quadratic function?
5. What is the significance of critical points in the context of optimization problems?
6. How would you solve a polynomial equation if it couldn't be factored?
7. What role do inflection points play in the analysis of functions?
8. How do the first and second derivatives together describe the shape of a graph?

**Tip:** When analyzing the nature of critical points, the sign of the second derivative is key: a positive value indicates a local minimum, and a negative value indicates a local maximum.

Learn how to identify critical points for the function f(x) = 3x^4 - 16x^3 + 6x^2 + 72x - 7 with detailed steps and analysis. Explore the first and second derivatives, cubic equations, and the second derivative test to classify local extrema. Suitable for advanced high school students (Grades 11-12).

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