Let f be a function such that f (1) = − 2 and f (5) = 7. Which of the following conditions ensures that f(c) = 0 for some value c in the open interval (1,5)?Responses∫15𝑓(𝑥)𝑑𝑥 exists. Integration of f of x of d x exists from 1 to 5f is increasing on the closed interval [1, 5].f is increasing on the closed interval [1, 5].f is continuous on the closed interval [1, 5].f is continuous on the closed interval [1, 5].f is defined for all values of x in the closed interval [1, 5].

Determine whether the statement is true or false.If f(1) > 0 and f(7) < 0, then there exists a number c between 1 and 7 such that f(c) = 0.TrueFalse

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.Tutorial ExerciseFor what value of the constant c is the function f continuous on (−∞, ∞)?f(x) =   cx2 + 5x      if x < 6x3 − cx if x ≥ 6Part 1 of 2Note that f is continuous on (−∞, 6) and (6, ∞). For the function to be continuous on (−∞, ∞), we need to ensure that as x approaches 6, the left and right limits match.First we find the left limit.lim x→6− f(x) = lim x→6− (cx2 + 5x) = Next, we find the right limit.lim x→6+ f(x) = lim x→6+ (x3 − cx) =

Verify that the function satisfies the three hypotheses of Rolle's theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle's theorem. (Enter your answers as a comma-separated list.)f(x) = 3x2 − 6x + 4,   [−1, 3]

Let ff be a continuous function such that integral, from, 1, to, 6, of, f, of, x, d, x, equals, 2∫ 16​ f(x)dx=2 and integral, from, 6, to, minus, 6, of, f, of, x, d, x, equals, 8, .∫ 6−6​ f(x)dx=8. What is the value of integral, from, minus, 6, to, 1, of, f, of, x, d, x, question mark∫ −61​ f(x)dx? ​

Let f:→R→(0,∞) be strictly increasing function such that limx→∞f(7x)f(x)=1. Then, the value of limx→∞[f(5x)f(x)−1] is equal to