Understanding the Intermediate Value Theorem with an Example

Intermediate Value Theorem (IVT)

The Intermediate Value Theorem (IVT) is a fundamental concept in calculus that deals with the continuity of functions

The Intermediate Value Theorem (IVT) is a fundamental concept in calculus that deals with the continuity of functions. The theorem states that if a function f(x) is continuous over a closed interval [a, b], and if k is any value between f(a) and f(b), then there exists at least one value c in the interval [a, b] such that f(c) = k.

In simpler terms, the IVT tells us that if a continuous function takes on two different values at the endpoints of an interval, then it must also take on every other value in between those two values at some point within the interval.

Let’s understand this concept through an example:

Consider the function f(x) = x^2 – 4x + 3 over the interval [1, 3].

Step 1: Check continuity
First, we need to ensure that the function is continuous over the interval [1, 3]. The function f(x) is a polynomial, and polynomials are continuous everywhere. Thus, f(x) is continuous over [1, 3].

Step 2: Determine f(a) and f(b)
Next, we evaluate the function at the interval endpoints. Let a = 1 and b = 3.
f(1) = 1^2 – 4(1) + 3 = 0
f(3) = 3^2 – 4(3) + 3 = 0

Step 3: Choose an intermediate value k
Choose any value k between f(a) and f(b). Since both f(1) and f(3) are 0, any value between them will satisfy the condition of the IVT. For simplicity, let’s choose k = 1.

Step 4: Verify the existence of c
Now, we need to show that there exists a value c in the interval [1, 3] such that f(c) = k = 1. This can be done by finding a specific x-value that satisfies f(x) = 1.

Let’s solve the equation:
x^2 – 4x + 3 = 1
Rearranging terms:
x^2 – 4x + 2 = 0

Using the quadratic formula:
x = (-(-4) ± sqrt((-4)^2 – 4(1)(2)))/(2(1))
x = (4 ± sqrt(16 – 8))/(2)
x = (4 ± sqrt(8))/(2)
x = (4 ± 2sqrt(2))/(2)
x = 2 ± sqrt(2)

Since x = 2 ± sqrt(2), which falls within the interval [1, 3], we can conclude that there exists at least one value c in the interval [1, 3] such that f(c) = 1.

Therefore, based on the Intermediate Value Theorem, we can conclude that the function f(x) = x^2 – 4x + 3 takes on the value 1 at some point within the interval [1, 3].

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