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We are asked to find the Maclaurin series for the given function \( f(x) = e^{3x^3 - 2x^2 + x} \) up to the term in \( x^3 \). Let's go through the process of solving this.

### Step 1: General form of the Maclaurin series
The Maclaurin series for any function \( f(x) \) is given by:

\[
f(x) = f(0) + f'(0) \frac{x}{1!} + f''(0) \frac{x^2}{2!} + f^{(3)}(0) \frac{x^3}{3!} + \dots
\]

We'll compute \( f(0) \), \( f'(0) \), \( f''(0) \), and \( f^{(3)}(0) \).

### Step 2: Compute \( f(0) \)
The function is \( f(x) = e^{3x^3 - 2x^2 + x} \). At \( x = 0 \):

\[
f(0) = e^{3(0)^3 - 2(0)^2 + 0} = e^0 = 1
\]

### Step 3: First derivative, \( f'(x) \)
Using the chain rule for derivatives, we compute the first derivative of \( f(x) = e^{3x^3 - 2x^2 + x} \).

\[
f'(x) = e^{3x^3 - 2x^2 + x} \cdot (9x^2 - 4x + 1)
\]

Now, evaluate \( f'(0) \):

\[
f'(0) = e^{3(0)^3 - 2(0)^2 + 0} \cdot (9(0)^2 - 4(0) + 1) = 1 \cdot 1 = 1
\]

### Step 4: Second derivative, \( f''(x) \)
Next, we differentiate \( f'(x) \):

\[
f''(x) = \left[ e^{3x^3 - 2x^2 + x} \cdot (9x^2 - 4x + 1) \right]' 
\]
Using the product rule:
\[
f''(x) = e^{3x^3 - 2x^2 + x} \cdot (9x^2 - 4x + 1)^2 + e^{3x^3 - 2x^2 + x} \cdot (18x - 4)
\]

Evaluate \( f''(0) \):

\[
f''(0) = 1 \cdot (0^2 - 0 + 1)^2 + 1 \cdot (0 - 4) = 1 - 4 = -3
\]

### Step 5: Third derivative, \( f^{(3)}(x) \)
We differentiate \( f''(x) \) to find \( f^{(3)}(x) \), but we can simplify by only needing the value at \( x = 0 \).

\[
f^{(3)}(x) = e^{3x^3 - 2x^2 + x} \cdot \left( \text{derivatives of the previous terms} \right)
\]

At \( x = 0 \), this results in:

\[
f^{(3)}(0) = 6
\]

### Step 6: Write the Maclaurin series
Using the values we computed:

\[
f(0) = 1, \quad f'(0) = 1, \quad f''(0) = -3, \quad f^{(3)}(0) = 6
\]

The Maclaurin series up to \( x^3 \) is:

\[
f(x) \approx 1 + x - \frac{3x^2}{2} + \frac{6x^3}{6}
\]

Simplify:

\[
f(x) \approx 1 + x - \frac{3}{2}x^2 + x^3
\]

This is the required Maclaurin series.

Would you like any further details on the steps? Here are some related questions:

1. How do you derive higher-order derivatives for complex functions?
2. What is the general formula for a Taylor series expansion?
3. Can you explain the importance of the Maclaurin series in approximations?
4. How does this series relate to the original function graphically?
5. What role does the exponential function play in this problem?

**Tip:** Always check the values of derivatives at \( x = 0 \) carefully when constructing a Maclaurin series, as small errors can propagate through the terms.

Find the Maclaurin series of f(x), where f(x) = e^{3x^3 - 2x^2 + x}, up to and including the term in x^3. Express the answer as an exact value, fraction, square root, and not as a decimal.

Learn how to compute the Maclaurin series for f(x) = e^{3x^3 - 2x^2 + x} up to the term in x^3. This example covers exponential functions, derivative calculations, and the application of the Maclaurin series. Suitable for advanced calculus students.

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