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Intermediate Value Theorem (IVT)

The Intermediate Value Theorem (IVT) is a fundamental theorem in calculus that helps us understand the behavior of continuous functions. It essentially states that if a continuous function takes on two values, then it must also take on all the values in between.

Statement of the IVT

Let $ f $ be a continuous function on the closed interval $ [a,b]] $ . If $ f(a) \neq f(b) $ , then for any value $ k $ between $ f(a) $ and $ f(b) $ , there exists a value $ c $ in the interval $ (a,b) $ such that $ f(c) = k $ .

Visualizing the IVT

Imagine a continuous curve representing the function $ f(x) $ on the interval $ [a,b]] $ . The IVT says that if the curve starts at a point with a certain y-value and ends at a point with a different y-value, then it must pass through every y-value in between those two points.

Example: Consider the function $ f(x) = x^2 $ on the interval $ [0,2 $ . $ f(0) = 0 $ and $ f(2) = 4 $ . Since $ f(x) $ is continuous, the IVT tells us that for any value $ k $ between 0 and 4, there exists a value $ c $ in the interval $ (0,2) $ such that $ f(c) = k $ .

Desmos Exploration

Key Points

• Continuity is crucial: The IVT only applies to continuous functions. If a function has a jump, hole, or vertical asymptote within the interval, the theorem may not hold.
• Closed interval: The IVT requires the interval to be closed, meaning it includes both endpoints.
• Intermediate value: The theorem guarantees the existence of a value $ c $ where $ f(c) = k $ , but it doesn’t tell us how to find that value.

Applications of the IVT

The IVT has numerous applications in calculus and other areas of mathematics. Some common uses include:

• Finding roots: The IVT can be used to prove that a function has a root (a value where the function equals zero) within a given interval.
• Solving equations: The IVT can help us determine if an equation has a solution within a specific interval.
• Proving inequalities: The IVT can be used to prove inequalities by showing that a function takes on a specific value within an interval.

Example: Proving the Existence of a Root

Let $ f(x) = x^3 - 2x - 5 $ . We want to prove that $ f(x) $ has a root in the interval $ 2,3 $ .

1. Continuity: $ f(x) $ is a polynomial function, which is continuous everywhere.
2. Interval: We are considering the closed interval $ 2,3 $ .
3. Values at endpoints: $ f(2) = 8 - 4 - 5 = -1 $ and $ f(3) = 27 - 6 - 5 = 16 $ .
4. Intermediate value: Since $ f(2) $ is negative and $ f(3) $ is positive, the IVT guarantees that there exists a value $ c $ in the interval $ (2,3) $ such that $ f(c) = 0 $ .

Therefore, we have proven that $ f(x) = x^3 - 2x - 5 $ has a root in the interval $ 2,3 $ .

Note: The IVT only proves the existence of a root, not its exact value. To find the root, we would need to use numerical methods like the Bisection Method.