Final answer:
To prove that the function g(x) = √f(x) is regulated, we need to show that it satisfies the definition of regulated functions. A function f(x) is regulated if it is bounded and has a finite number of discontinuities on the interval [a, b].
Step-by-step explanation:
To prove that the function g(x) = √f(x) is regulated, we need to show that it satisfies the definition of regulated functions. A function f(x) is regulated if it is bounded and has a finite number of discontinuities on the interval [a, b].
First, notice that since f(x) is non-negative for all x in [a, b], the function g(x) = √f(x) is also non-negative. This means that g(x) is bounded below by 0. We need to prove that g(x) is also bounded above. Since f(x) is regulated, it is bounded, which means there exist constants M and N such that 0 ≤ M ≤ f(x) ≤ N for all x in [a, b]. Taking the square root of this inequality, we get 0 ≤ √M ≤ √f(x) ≤ √N for all x in [a, b]. Therefore, g(x) = √f(x) is bounded by √M and √N for all x in [a, b].
Next, let's consider the discontinuities. Suppose there is a point c in [a, b] where f(x) has a discontinuity. This means that there exist two sequences (x_n) and (y_n) in [a, b] that approach c from the left and right, respectively, such that lim (x_n) = lim (y_n) = c and lim f(x_n) ≠ lim f(y_n). Taking the square root of both sides of the inequality, we have lim √f(x_n) ≠ lim √f(y_n). Therefore, g(x) = √f(x) has a discontinuity at c. Since there are only a finite number of discontinuities in f(x), there will also be a finite number of discontinuities in g(x).