To find the range of values of $x$ for which the function $f(x) = e^x \cos x$ is increasing, we need to determine where the derivative of the function is positive, i.e., $f'(x) > 0$.

Step 1: Compute the derivative of $f(x) = e^x \cos x$

Using the product rule for differentiation, we have:

$f'(x) = \frac{d}{dx}(e^x \cos x) = e^x \cdot \frac{d}{dx}(\cos x) + \cos x \cdot \frac{d}{dx}(e^x)$

$f'(x) = e^x (-\sin x) + e^x \cos x$

Thus,

$f'(x) = e^x (\cos x - \sin x)$

Step 2: Set the derivative greater than zero

For the function to be increasing, we require:

$f'(x) > 0$

This leads to:

$e^x (\cos x - \sin x) > 0$

Since $e^x > 0$ for all real $x$, the inequality reduces to:

$\cos x - \sin x > 0$

Step 3: Solve $\cos x - \sin x > 0$

Dividing the inequality by $\sqrt{2}$, we have:

$\frac{\cos x}{\sqrt{2}} - \frac{\sin x}{\sqrt{2}} > 0$

This simplifies to:

$\cos \left( x + \frac{\pi}{4} \right) > 0$

The cosine function is positive when the angle lies between:

$- \frac{\pi}{2} + 2n\pi < x + \frac{\pi}{4} < \frac{\pi}{2} + 2n\pi$

Step 4: Solve for $x$

Subtract $\frac{\pi}{4}$ from all sides:

$- \frac{\pi}{2} - \frac{\pi}{4} + 2n\pi < x < \frac{\pi}{2} - \frac{\pi}{4} + 2n\pi$

$- \frac{3\pi}{4} + 2n\pi < x < \frac{\pi}{4} + 2n\pi$

Thus, $x$ lies in the intervals:

$x \in \left( -\frac{3\pi}{4} + 2n\pi, \frac{\pi}{4} + 2n\pi \right)$

where $n \in \mathbb{Z}$.

Final Answer

The function $f(x) = e^x \cos x$ is increasing in the intervals:

$x \in \left( -\frac{3\pi}{4} + 2n\pi, \frac{\pi}{4} + 2n\pi \right) \quad \text{for} \quad n \in \mathbb{Z}$

Would you like further details or have any questions?

Here are some related questions to explore:

1. What is the decreasing interval for $f(x) = e^x \cos x$?
2. How does the function behave at critical points?
3. What is the maximum value of $f(x)$ on a given interval?
4. How do the increasing and decreasing intervals of $e^x \sin x$ compare to $e^x \cos x$?
5. How would the range change if we considered $e^{-x} \cos x$?

Tip: Always check the behavior of trigonometric functions over their periodic intervals when solving inequalities like these!