According to the intermediate value theorem, if you have a function where f(4) = 5 and f(6) = 3, there will be at least one point x between 4 and 6 where f(x) = 4.

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.x4 + x − 4 = 0,    (1, 2)f(x) = x4 + x − 4 is on the closed interval [1, 2], f(1) = , and f(2) = .

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.x4 + x − 5 = 0,    (1, 2)f(x) = x4 + x − 5 is on the closed interval [1, 2], f(1) = , and f(2) = . Since −3 < < 13, there is a number c in (1, 2) such that f(c) = by the Intermediate Value Theorem. Thus, there is a of the equation x4 + x − 5 = 0 in the interval (1, 2).

Let f(x) = −x5 −2x4 + 12x3 +x−1. Then either show that there is an x such thatf(x) = 0, and find an interval where it exists; or, show there is no such x.

We cannot algebraically solve 2x = 3x but we can locate the solutions using the Intermediate Value Theorem. Consider the continuous function f(x) = 2x − 3x. The roots of are exactly the solutions to our equation. Consider the following intervals.(0, 1)(1, 2)(2, 3)(3, 4)On which of the above intervals, does the Intermediate Value Theorem guarantee a solution to the equation 2x = 3x?1 and 42 and 4    1, 2, and 41 and 3

Suppose that 4 ≤ f ′(x) ≤ 5 for all values of x. What are the minimum and maximum possible values of f(6) − f(3)? ≤ f(6) − f(3)