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Find the extreme value for the following functions f(x)=2x³-2x²-36x+3

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The function you want to find the extreme values for is:
\[ f(x) = 2x^3 - 2x^2 - 36x + 3 \]

To find the extreme values, we need to follow these steps:

1. **Find the first derivative** \( f'(x) \).
2. **Set** \( f'(x) = 0 \) **to find critical points**.
3. **Determine the nature of the critical points** using the second derivative test or by analyzing the sign changes in the first derivative.

### Step 1: Find the first derivative
\[ f'(x) = \frac{d}{dx}(2x^3 - 2x^2 - 36x + 3) \]

### Step 2: Solve \( f'(x) = 0 \) to find critical points

### Step 3: Determine the nature of the critical points

Let me calculate this for you.### Step-by-Step Solution

1. **First Derivative**:
   The first derivative of the function \( f(x) = 2x^3 - 2x^2 - 36x + 3 \) is:
   \[
   f'(x) = 6x^2 - 4x - 36
   \]

2. **Critical Points**:
   We solve \( f'(x) = 0 \) to find the critical points:
   \[
   6x^2 - 4x - 36 = 0
   \]
   The solutions (critical points) are:
   \[
   x = \frac{1 - \sqrt{55}}{3} \quad \text{and} \quad x = \frac{1 + \sqrt{55}}{3}
   \]

3. **Second Derivative Test**:
   The second derivative is:
   \[
   f''(x) = 12x - 4
   \]
   Substituting the critical points into the second derivative:
   - For \( x = \frac{1 - \sqrt{55}}{3} \), \( f''(x) = -4\sqrt{55} \) (negative), indicating a local maximum.
   - For \( x = \frac{1 + \sqrt{55}}{3} \), \( f''(x) = 4\sqrt{55} \) (positive), indicating a local minimum.

### Conclusion
The function \( f(x) = 2x^3 - 2x^2 - 36x + 3 \) has:
- A **local maximum** at \( x = \frac{1 - \sqrt{55}}{3} \).
- A **local minimum** at \( x = \frac{1 + \sqrt{55}}{3} \).

Would you like further details or assistance with other questions? 

### Relative Questions:
1. How do you calculate inflection points for a cubic function?
2. Can you explain the second derivative test in more detail?
3. What are the possible types of critical points in a function?
4. How do we interpret the results of the second derivative test?
5. What is the significance of the discriminant in solving quadratic equations?

### Tip:
Remember, the sign of the second derivative at a critical point helps determine the concavity of the function, which in turn reveals whether the point is a maximum, minimum, or a saddle point.

Learn how to find extreme values for the function f(x) = 2x³ - 2x² - 36x + 3 using calculus. Understand the process of finding critical points, determining local maxima and minima, and applying the second derivative test. Suitable for advanced high school students and anyone interested in calculus.

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