*f*(*x*)=2*ax*3−*ax*2+2

*a*>0


Calculate the extreme points, if necessary use parameter a

Given the function \( f(x) = 2ax^3 - ax^2 + 2 \), where \( a > 0 \), we need to find the extreme points by taking the derivative and finding where it equals zero.

### Step 1: Find the first derivative of \( f(x) \).
The first derivative of \( f(x) \) with respect to \( x \) is:

\[
f'(x) = \frac{d}{dx} \left( 2ax^3 - ax^2 + 2 \right)
\]

Differentiate each term:

\[
f'(x) = 6ax^2 - 2ax
\]

### Step 2: Set the first derivative equal to zero to find critical points.
To find the critical points, solve the equation \( f'(x) = 0 \):

\[
6ax^2 - 2ax = 0
\]

Factor out \( 2ax \) from the equation:

\[
2ax(3x - 1) = 0
\]

This gives two solutions:

1. \( 2ax = 0 \) \(\Rightarrow\) \( x = 0 \)
2. \( 3x - 1 = 0 \) \(\Rightarrow\) \( x = \frac{1}{3} \)

### Step 3: Determine the nature of these critical points.
To determine whether these critical points are maxima, minima, or points of inflection, we can use the second derivative test.

### Step 4: Find the second derivative of \( f(x) \).
The second derivative of \( f(x) \) is:

\[
f''(x) = \frac{d}{dx} \left( 6ax^2 - 2ax \right) = 12ax - 2a
\]

### Step 5: Evaluate the second derivative at each critical point.

1. At \( x = 0 \):

\[
f''(0) = 12a(0) - 2a = -2a
\]

Since \( a > 0 \), \( f''(0) = -2a < 0 \), indicating that \( x = 0 \) is a **local maximum**.

2. At \( x = \frac{1}{3} \):

\[
f''\left(\frac{1}{3}\right) = 12a\left(\frac{1}{3}\right) - 2a = 4a - 2a = 2a
\]

Since \( 2a > 0 \), \( f''\left(\frac{1}{3}\right) > 0 \), indicating that \( x = \frac{1}{3} \) is a **local minimum**.

### Step 6: Find the coordinates of the extreme points.
Finally, substitute the values of \( x \) back into the original function \( f(x) \) to find the corresponding \( y \)-coordinates.

1. For \( x = 0 \):

\[
f(0) = 2a(0)^3 - a(0)^2 + 2 = 2
\]

So, the local maximum is at \( (0, 2) \).

2. For \( x = \frac{1}{3} \):

\[
f\left(\frac{1}{3}\right) = 2a\left(\frac{1}{3}\right)^3 - a\left(\frac{1}{3}\right)^2 + 2 = 2a\left(\frac{1}{27}\right) - a\left(\frac{1}{9}\right) + 2 = \frac{2a}{27} - \frac{a}{9} + 2
\]

Simplifying further:

\[
f\left(\frac{1}{3}\right) = \frac{2a}{27} - \frac{3a}{27} + 2 = \frac{-a}{27} + 2
\]

Thus, the local minimum is at \( \left(\frac{1}{3}, 2 - \frac{a}{27}\right) \).

### Conclusion:
The extreme points are:
- **Local Maximum:** \( (0, 2) \)
- **Local Minimum:** \( \left(\frac{1}{3}, 2 - \frac{a}{27}\right) \)

Would you like more details on this process, or have any other questions?

Here are 5 related questions you might consider:
1. How do changes in the parameter \( a \) affect the location of the extreme points?
2. Can you explain the role of the second derivative in determining the concavity of the function?
3. How would the extreme points change if \( a < 0 \)?
4. What does the behavior of the function at infinity tell us about the graph?
5. How can we apply these concepts to optimize real-world problems?

**Tip:** Always check the sign of the second derivative at critical points to confirm whether they are maxima or minima.

Learn how to find extreme points of the function f(x) = 2ax^3 - ax^2 + 2 with a detailed step-by-step solution. This problem involves calculus, derivatives, and the second derivative test, applicable to advanced high school students and beyond.

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