Question

Suppose that f is defined in a neighborhood of x₀ and that f is continuous at x₀ . And suppose that f(x₀ )>0. Show that there exists a neighborhood of xV where f is positive.

Answer 1

To demonstrate the existence of a positive-value neighborhood for a continuous function at a point, we use the definition of continuity which ensures that within a certain range around that point, the function's values will not deviate beyond a chosen positive bound from the function's value at that point.

Step-by-step explanation:

The question asks us to prove that if a function f is continuous at a point x₀ and f(x₀) is positive, then there is a neighborhood around x₀ where f remains positive. This can be shown by using the definition of continuity. Since f is continuous at x₀, for any ε greater than zero, there exists a δ such that for all x in the neighborhood (x₀-δ, x₀+δ), the value of f(x) is within ε of f(x₀). Choosing ε to be f(x₀) itself ensures that f(x) stays positive within this neighborhood, as f(x) cannot be lower than zero (as that would violate the ε criterion).

Continuing this line of reasoning, we can see that the concept of a continuous probability function is crucial in understanding probability as an area under the probability density function f(x). In any continuous probability distribution, the probability is reflected as the area under the curve of the distribution, between the specified values of x.