To find the critical points of the function $g(x) = e^x \cos x$ on the interval $[0, 2\pi]$, we need to follow these steps:

Step 1: Differentiate $g(x)$

The critical points occur where the derivative of the function equals zero or does not exist. We first compute the derivative $g'(x)$.

Using the product rule for derivatives: $g'(x) = \frac{d}{dx} (e^x \cos x) = \frac{d}{dx} (e^x) \cdot \cos x + e^x \cdot \frac{d}{dx} (\cos x)$ $g'(x) = e^x \cos x - e^x \sin x$ So, the derivative is: $g'(x) = e^x (\cos x - \sin x)$

Step 2: Solve $g'(x) = 0$

To find the critical points, we solve: $e^x (\cos x - \sin x) = 0$ Since $e^x$ is never zero for any real $x$, we need to solve: $\cos x - \sin x = 0$ This simplifies to: $\cos x = \sin x$ Dividing both sides by $\cos x$ (assuming $\cos x \neq 0$): $1 = \tan x$ Thus, $x = \frac{\pi}{4} + n\pi$ for some integer $n$.

Step 3: Find solutions in the interval $[0, 2\pi]$

Now we find the values of $x$ within the interval $[0, 2\pi]$: $x = \frac{\pi}{4}, \frac{5\pi}{4}$

Step 4: Check the boundaries

Since the interval is closed, we also check the boundaries $x = 0$ and $x = 2\pi$. However, $g'(0)$ and $g'(2\pi)$ do not result in zeros, so they are not critical points.

Conclusion:

The critical points of $g(x) = e^x \cos x$ on the interval $[0, 2\pi]$ are: $x = \frac{\pi}{4}, \frac{5\pi}{4}$

Would you like further clarification or details on any step?

Here are 5 related questions for further exploration:

1. How would the critical points change if we considered the interval $[-2\pi, 2\pi]$?
2. What is the behavior of the function $g(x) = e^x \cos x$ near its critical points?
3. Can you find the inflection points of $g(x)$ by taking the second derivative?
4. What happens to the function as $x \to \infty$?
5. How would you use the second derivative test to determine if the critical points are maxima or minima?

Tip:

The product rule and chain rule are essential tools in differentiating composite functions, so make sure you are comfortable with both.