Verify that the Intermediate Value theorem applies to the indicated interval and find the value of c guaranteed by the theorem.
f(x) = x^2 - 6x + 8, [0,3], f(c) = 0
I have no idea how to use the theorem :(
No problem! I'll explain how to use the Intermediate Value Theorem in this case.
The Intermediate Value Theorem states that if a function is continuous on a closed interval [a, b], and if f(a) and f(b) are of opposite signs, then there exists at least one number c in the interval (a, b) such that f(c) = 0.
In this problem, you're given the function f(x) = x^2 - 6x + 8 and the interval [0,3]. You're also asked to find the value of c guaranteed by the theorem, such that f(c) = 0.
To verify if the Intermediate Value Theorem applies to the given interval [0, 3], you need to check if f(0) and f(3) have opposite signs. If they do, then the theorem applies.
Let's calculate f(0) and f(3) to check their signs:
f(0) = (0)^2 - 6(0) + 8 = 8
f(3) = (3)^2 - 6(3) + 8 = 1
Since f(0) = 8 and f(3) = 1, notice that the signs are different. f(0) is positive and f(3) is positive. This means that the Intermediate Value Theorem applies to the interval [0, 3].
Now, we need to find the value of c within the interval (0, 3) such that f(c) = 0. Since f(x) is a continuous function within the interval [0, 3] and f(0) and f(3) have different signs, there must be at least one value of c between 0 and 3 where f(c) = 0.
To find the specific value of c, you can use various numerical methods such as the Bisection Method, Newton's Method, or graphical methods. However, for this particular problem, let's solve the equation f(x) = 0 algebraically.
x^2 - 6x + 8 = 0
This quadratic equation can be factored as follows:
(x - 4)(x - 2) = 0
Setting each factor equal to zero, we have:
x - 4 = 0 or x - 2 = 0
x = 4 or x = 2
Therefore, the Intermediate Value Theorem guarantees the existence of a value c within the interval [0, 3] such that f(c) = 0. In this case, the values of c are 4 and 2.