COURSE: Mathematical Analysis/Real Analysis (CC1A)TOPIC: Continuity + Connectedness **Prove \( f(x) \) has a zero (i.e., a point where \( f(p) = 0 \)) on each interval. You can assert that the functions are continuous on the relevant intervals (i.e., a rigorous proof that \( f(x) \) is continuous on the given interval is not needed).**\[ f(x) = 2e^{-x} (\cos(2x)); \text{ on } \left[0, \frac{\pi}{2}\right] \]---### Explanation:The function \( f(x) = 2e^{-x} (\cos(2x)) \) is defined on the interval \([0, \frac{\pi}{2}]\). The goal is to demonstrate that a zero exists within this interval, meaning there exists a point \( p \) within \([0, \frac{\pi}{2}]\) where \( f(p) = 0 \). Continuity within the interval can be assumed without a detailed proof due to the nature of exponential and trigonometric functions composing \( f(x) \).

We are given a function f(x) = 2e-x (cos(2x)). We need to prove that f(x) has a zero on each interval. To do so, we need to use the following concept.Intermediate Value Theorem:Let f be a function f: ℝ→ℝ which is continuous on interval [a, b]. If f(a) < f(b) and f(a) < c < f(b); then there exists a point x∈(a, b) such that f(x) = c.Now, note that the functions g(x) = 2e-x and h(x) = cos(2x) , are continuous over entire domain ℝ. We know that the product of two continuous functions are continuous, provided their domain and range are same. Thus, the function f(x) = 2e-x (cos(2x)) is continuous over all ℝ. Now, let [a, b] be any interval in ℝ, such that f(a) < 0 and f(b) > 0. Then, we have the following points:1. f is continuous on [a, b] ,2. Since, f(a) < 0 and f(b) > 0, we have f(a) < f(b) 3. Let c = 0, then f(a) < 0 < f(b), Then, by intermediate mean value. ∃ x∈[a, b] , such that f(x) = 0. Thus, for every relevant interval [a, b], f(x) has a zero.