To solve the given problem, we will:

1. Find the critical points by calculating the derivative $f'(x)$ and setting it equal to zero.
2. Classify the critical points using the second derivative $f''(x)$ to determine whether they correspond to local maxima, minima, or points of inflection.
3. Analyze the behavior of $f(x)$ as $x \to \pm \infty$.

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Step 1: Finding the critical points

The function is given as:

$f(x) = x^3 - 3x^2 + 4 \sin(x) e^x$

Now, we calculate the first derivative:

$f'(x) = \frac{d}{dx}\left(x^3 - 3x^2 + 4 \sin(x) e^x\right)$

Using the product rule for the term $4 \sin(x) e^x$:

$f'(x) = 3x^2 - 6x + 4 \left(\sin(x) e^x + \cos(x) e^x \right)$

Simplifying:

$f'(x) = 3x^2 - 6x + 4e^x (\sin(x) + \cos(x))$

To find the critical points, set $f'(x) = 0$:

$3x^2 - 6x + 4e^x (\sin(x) + \cos(x)) = 0$

This equation must be solved for $x$. Unfortunately, it’s a transcendental equation due to the $e^x$ term, so exact solutions might not be expressible in closed form. You can approximate the solutions numerically, but for now, we'll move to the next step of analyzing the nature of critical points in general.

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Step 2: Classifying the critical points (Local Maxima/Minima)

To classify the critical points, we compute the second derivative $f''(x)$:

$f''(x) = \frac{d}{dx} \left(3x^2 - 6x + 4e^x (\sin(x) + \cos(x))\right)$

This gives:

$f''(x) = 6x - 6 + 4e^x\left((\sin(x) + \cos(x)) + (\cos(x) - \sin(x))\right)$

Simplifying the expression:

$f''(x) = 6x - 6 + 4e^x(2\cos(x))$

Now, at any critical point $x = c$, we evaluate $f''(c)$:

• If $f''(c) > 0$, the point $x = c$ is a local minimum.
• If $f''(c) < 0$, the point $x = c$ is a local maximum.
• If $f''(c) = 0$, the second derivative test is inconclusive, and further analysis is needed.

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Step 3: Behavior of $f(x)$ as $x \to \pm \infty$

• As $x \to \infty$, the term $x^3$ dominates, and since $x^3 \to \infty$, the function $f(x) \to \infty$.
• As $x \to -\infty$, the term $x^3$ dominates again, and since $x^3 \to -\infty$, the function $f(x) \to -\infty$.

Thus, the behavior of $f(x)$ at infinity is:

$\lim_{x \to \infty} f(x) = \infty, \quad \lim_{x \to -\infty} f(x) = -\infty$

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Conclusion:

1. The critical points are found by solving $f'(x) = 0$, which involves solving the transcendental equation.
2. These critical points can be classified by using the second derivative $f''(x)$.
3. As $x \to \infty$, $f(x) \to \infty$, and as $x \to -\infty$, $f(x) \to -\infty$.

Would you like me to help further with numerical approximations for the critical points?

Follow-up questions:

1. How do you solve transcendental equations numerically?
2. What does the second derivative test tell you about the curvature of a function?
3. How do local maxima differ from global maxima?
4. What is the significance of the behavior of $f(x)$ as $x \to \infty$?
5. How does the combination of trigonometric and exponential terms influence the behavior of $f(x)$?

Tip:

When analyzing the behavior at infinity, focus on the highest-degree polynomial term to determine the overall trend.