Analyse the function f(x) = e^(-alpha*x) * cos(x): find its value at 0, the period of its oscillation, its envelope curves, and how the sign of alpha changes its behaviour.

Step-by-step solution

1. 1

   Split the function into an amplitude factor and an oscillating factor. Compare with the general damped oscillation $Ae^{-\alpha x}\cos(kx + \phi)$: here $A = 1$, $k = 1$ and $\phi = 0$. The exponential controls how big the swings are; the cosine controls how fast they happen. Neither factor affects the other's role.

2. 2

   Evaluate at the origin.

   $f(0) = e^{0}\cos(0) = 1 \times 1 = 1$

   So the curve starts at its maximum possible amplitude for any $\alpha$.

3. 3

   Find the period of the oscillation. The zeros and turning-point pattern repeat whenever $\cos x$ does, so

   $T = \frac{2\pi}{k} = \frac{2\pi}{1} = 2\pi$

   Strictly the function itself is not periodic unless $\alpha = 0$ — the pattern repeats but the size does not.

4. 4

   Derive the envelope curves. Since $-1 \le \cos x \le 1$ and $e^{-\alpha x} > 0$, multiplying through preserves the inequality:

   $-e^{-\alpha x} \le e^{-\alpha x}\cos x \le e^{-\alpha x}$

   so the graph is trapped between $y = e^{-\alpha x}$ and $y = -e^{-\alpha x}$, touching each alternately at $x = n\pi$.

5. 5

   Split on the sign of α. If $\alpha > 0$ then $e^{-\alpha x} \to 0$ as $x \to \infty$ and the oscillations decay to zero — the underdamped regime. If $\alpha = 0$ the function is exactly $\cos x$, undamped and genuinely periodic. If $\alpha < 0$ the envelope grows and the oscillations blow up.

6. 6

   Check the envelope contact points. At $x = 0$: $f = 1$ and the upper envelope is $e^0 = 1$ — they touch. At $x = \pi$: $\cos\pi = -1$, so $f(\pi) = -e^{-\alpha\pi}$, touching the lower envelope. For $\alpha = 0.2$ this gives $f(\pi) = -0.533488$, and the successive extrema shrink by the constant ratio $e^{-0.2\pi} = 0.533488$ each half-period — the hallmark of exponential damping.