Question

Test your understanding of the Intermediate Value Theorem. Let \(f\) be continuous on [1,5] where \(f(1)=-2\) and \(f(5)=\) -10. Does a value \(1

Short answer

No, there is no such value \( 1 < c < 5 \) because \( f(1) = -2 \) and \( f(5) = -10 \), and \(-9\) is not between \(-2\) and \(-10\).

Step-by-step solution

Understanding the Intermediate Value Theorem

The Intermediate Value Theorem (IVT) states that if a function \( f \) is continuous on the interval \([a, b]\), and \( N \) is any number between \( f(a) \) and \( f(b) \), then there exists at least one number \( c \) in the interval \((a, b)\) such that \( f(c) = N \).

Identify the Given Values

We know that \( f \) is continuous on the interval \([1, 5]\), \( f(1) = -2 \), and \( f(5) = -10 \). The goal is to find a value between \( 1 \) and \( 5 \) where \( f(c) = -9 \).

Check if the Intermediate Value Theorem Applies

Since \( f(1) = -2 \) and \( f(5) = -10 \), the values of \( f(x) \) at the endpoints are \(-2\) and \(-10\). We are looking for \( f(c) = -9 \), which must lie between \( -2 \) and \( -10 \). However, there is no value between \(-2\) and \(-10\) satisfying \( f(c) = -9 \) because \( -9\) is not between \(-2\) and \(-10\).

Conclusion Based on the Intermediate Value Theorem

Since \(-9\) is not between \(-2\) and \(-10\), the Intermediate Value Theorem does not guarantee the existence of such a \( c \) on the interval \((1, 5)\).

Key concepts

Continuous Functions

Continuous functions are foundational to understanding many concepts in calculus, including the Intermediate Value Theorem. A function is considered continuous over an interval if there are no gaps, jumps, or breaks in the graph of that function within the interval. This means that as you move along the curve from one endpoint of the interval to the other, the function changes smoothly without any sudden jumps.

Key points to remember about continuous functions include:

• A continuous function on a closed interval equires that the function be defined at every point within that interval.
• The limit of the function at any point in the interval must equal the function's value at that point.

If a function is not continuous, it may not satisfy these conditions, and the Intermediate Value Theorem won't apply. For the given problem, the function is continuous over [1, 5], which is crucial because without continuity, we could not apply this theorem to assure the existence of a certain value between the endpoints of the interval.

Existence of a Value Within an Interval

One of the main ideas behind the Intermediate Value Theorem is the assurance of the existence of a value within an interval for a continuous function. The theorem tells us that between any two values that a continuous function takes on an interval, every intermediate value must also occur.

In the context of our problem, we are exploring whether there exists a value within the interval (1, 5) such that the function value is -9. Given that:

• \( f(1) = -2 \)
• \( f(5) = -10 \)

have been provided as endpoints, the theorem can be used to check for intermediate values between -2 and -10.

However, the number -9 does not lie between -2 and -10 on the number line, hence the Intermediate Value Theorem doesn’t assure us of a point \( c \) where \( f(c) = -9 \) within this interval. This tells us a critical limitation of the theorem: while it guarantees existence, it only applies to values that fall within the range of the endpoints.

Calculus Proofs

Calculus proofs often involve demonstrating why a certain mathematical statement or theorem holds true under specified conditions. The Intermediate Value Theorem offers a clear example of how proofs in calculus are structured.

For any proof involving the Intermediate Value Theorem, certain conditions need verification:

• The function must be continuous on the closed interval.
• The value in question must lie between the function values at the two endpoints.

In the solution to our problem, we followed the steps:1. Confirmed the function's continuity on the given interval [1, 5].2. Checked if -9 is an intermediate value between the function’s values at 1 and 5.

Deriving a calculus proof essentially means carefully assessing whether the premises of a theorem apply. In this case, since -9 was not between -2 and -10, the proof demonstrated that there is no guarantee of such \( c \) existing in the interval. Proof structures in calculus often follow a logical sequence of verifying conditions, applying the theorem, and then reaching a conclusion based on these verifications.