## Problem

Consider the polynomial function:

$$ f(x) = x^3 - 7x^2 + 14x - 8 $$

1. Find the first derivative of $ f(x) $.

2. Factor the derivative.

3. Use the factored derivative to determine the points where the slope of the function is zero (i.e., the critical points).


### 1. Find the first derivative of $ f(x) $:

Given:  
\[
f(x) = x^3 - 7x^2 + 14x - 8
\]  
Using basic differentiation rules:  
\[
f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(7x^2) + \frac{d}{dx}(14x) - \frac{d}{dx}(8)
\]

\[
f'(x) = 3x^2 - 14x + 14
\]

---

### 2. Factor the derivative:

We need to factor the quadratic polynomial $ f'(x) = 3x^2 - 14x + 14 $.  
Using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where $ a = 3 $, $ b = -14 $, and $ c = 14 $.

\[
x = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(3)(14)}}{2(3)}
= \frac{14 \pm \sqrt{196 - 168}}{6}
= \frac{14 \pm \sqrt{28}}{6}
\]

Since $ \sqrt{28} = 2\sqrt{7} $, the solutions are:

\[
x = \frac{14 \pm 2\sqrt{7}}{6} = \frac{7 \pm \sqrt{7}}{3}
\]

Thus, the roots are $ x_1 = \frac{7 + \sqrt{7}}{3} $ and $ x_2 = \frac{7 - \sqrt{7}}{3} $.

So, we can factor $ f'(x) $ as:
\[
f'(x) = 3 \left( x - \frac{7 + \sqrt{7}}{3} \right) \left( x - \frac{7 - \sqrt{7}}{3} \right)
\]

---

### 3. Find the critical points:

The critical points occur where $ f'(x) = 0 $, which are the solutions to the equation:

\[
3 \left( x - \frac{7 + \sqrt{7}}{3} \right) \left( x - \frac{7 - \sqrt{7}}{3} \right) = 0
\]

Thus, the critical points are:
\[
x = \frac{7 + \sqrt{7}}{3} \quad \text{and} \quad x = \frac{7 - \sqrt{7}}{3}
\]

---

### Summary:

1. **First derivative:**  
   \[ f'(x) = 3x^2 - 14x + 14 \]

2. **Factored derivative:**  
   \[ f'(x) = 3 \left( x - \frac{7 + \sqrt{7}}{3} \right) \left( x - \frac{7 - \sqrt{7}}{3} \right) \]

3. **Critical points:**  
   \[ x = \frac{7 + \sqrt{7}}{3}, \quad x = \frac{7 - \sqrt{7}}{3} \]

Would you like more details on how these points relate to maxima, minima, or concavity?  

---

Here are **5 related questions** to explore further:

1. What are the values of $ f(x) $ at the critical points?
2. How do we determine if these critical points are local maxima or minima?
3. What is the second derivative of $ f(x) $, and what does it tell us about concavity?
4. Can the derivative be factored differently, such as by completing the square?
5. How would the critical points change if we altered the coefficients of the original polynomial?

**Tip:** When checking for extrema, always use the second derivative test or analyze sign changes of the first derivative around the critical points.

Consider the polynomial function: f(x) = x^3 - 7x^2 + 14x - 8. 1. Find the first derivative of f(x). 2. Factor the derivative. 3. Use the factored derivative to determine the points where the slope of the function is zero (i.e., the critical points).

Explore how to find the critical points of the cubic function f(x) = x^3 - 7x^2 + 14x - 8 by taking the first derivative, factoring the quadratic, and solving for the points where the slope is zero. This problem covers differentiation and factoring techniques for Grades 10-12.

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