Determine the interval(s) on which the following function is continuous. Then analyze the given limits. 6 p X f(x): lim f(x); lim f(x) 1 - p X ' X-0 X-+0+ . . . The function is continuous on the interval(s) ( - co,0), (0,co) . (Simplify your answer. Type your answer in interval notation. Use a comma to separate answers as needed.) lim f(x) = X-0-a. Use the Intermediate Value Theorem to show that the following equation has a solution on the given interval. b. Use a graphing utility to find all the solutions to the equation on the given interval. c. Illustrate your answers with an appropriate graph. x' + x+ 3=0; (-2,1) a. The Intermediate Value Theorem states that if f is continuous on the interval [a,b] and L is a number strictly between f(a) and f(b), then there exists at least one number c in (a,b) satisfying f ( c) = L. For which values of x is the function f(x) = x + x + 3 continuous? A. It is continuous on [ - 2, 1], but not for all x. B. It is continuous for all x. C. It is continuous for some x, but not on [ - 2, 1]. O D. It is not continuous on any interval. Evaluate the function f(x) at the left endpoint. The value of the function at the left endpoint is - 7 . (Type an integer or decimal rounded to three decimal places as needed.) Evaluate the function f(x) at the right endpoint. The value of the function at the right endpoint is 5 . (Type an integer or decimal rounded to three decimal places as needed.) Why can the Intermediate Value Theorem be used to show that the equation has a solution on ( - 2, 1)? A. It can be used because f(x) = x + x + 3 is defined on ( - 2,1) and 0