Use The Intermediate Value Theorem To Determine The Interval Where The Equation Cos(x-5) = X-5 Has A Solution, Given The Intervals A. (3, 4), B. (4, 5), C. (5, 6), And D. (6, 7).

The Intermediate Value Theorem (IVT) is a fundamental concept in calculus that allows us to determine the existence of solutions to equations within specific intervals. In essence, it bridges the gap between the continuity of a function and the values it takes within a given range. This article will delve into the IVT, its applications, and how it can be used to pinpoint intervals where solutions to equations reside. We'll explore the equation $\cos(x-5) = x-5$ as a case study, demonstrating how the IVT helps us identify the correct interval containing its solution.

Understanding the Intermediate Value Theorem

At its core, the Intermediate Value Theorem is remarkably intuitive. Imagine a continuous curve drawn on a graph. If you pick two points on the curve, say A and B, and a y-value (let's call it 'k') that falls between the y-coordinates of A and B, then the IVT guarantees that there's at least one x-value between the x-coordinates of A and B where the curve crosses the horizontal line y = k. In simpler terms, a continuous function must take on every value between any two of its values. Let’s delve deeper into the mathematical formulation of this crucial theorem and explore its underlying principles.

The Intermediate Value Theorem states that if $f(x)$ is a continuous function on the closed interval $[a, b]$, and $k$ is any number between $f(a)$ and $f(b)$, then there exists at least one number $c$ in the interval $(a, b)$ such that $f(c) = k$. Continuity is a critical requirement for the IVT to hold. A function is continuous on an interval if its graph can be drawn without lifting the pen, meaning there are no breaks, jumps, or vertical asymptotes within that interval. Functions like polynomials, sine, cosine, and exponential functions are continuous over their domains. The value 'k' plays a pivotal role as it represents the intermediate value we are trying to find a corresponding 'c' for within the interval. If $f(a)$ and $f(b)$ have opposite signs (one positive and one negative), then 0 is an intermediate value between them. Consequently, the IVT guarantees the existence of a root (a solution where $f(c) = 0$) within the interval $(a, b)$.

To effectively apply the Intermediate Value Theorem, we need to verify that the function in question is indeed continuous on the given closed interval. This is usually straightforward for common functions like polynomials and trigonometric functions within their respective domains. Once continuity is established, the next step involves evaluating the function at the endpoints of the interval, denoted as $f(a)$ and $f(b)$. The crucial part is to check if the desired intermediate value 'k' lies between $f(a)$ and $f(b)$. In the context of finding solutions to equations, we often rearrange the equation into the form $f(x) = 0$, making 'k' equal to 0. Therefore, we look for a sign change between $f(a)$ and $f(b)$. If $f(a)$ and $f(b)$ have opposite signs, the IVT guarantees a root within the interval. If there's no sign change, the IVT doesn't guarantee a root, but it also doesn't rule one out – there might still be a solution within the interval. Consider a scenario where $f(a)$ and $f(b)$ are both positive, but the function dips below the x-axis within the interval and then rises back up. In such cases, there could be two roots within the interval, even though $f(a)$ and $f(b)$ are both positive.

Applying the IVT to $\cos(x-5) = x-5$

Now, let's apply the Intermediate Value Theorem to the equation $\cos(x-5) = x-5$. Our goal is to determine which of the given intervals, (3, 4), (4, 5), (5, 6), or (6, 7), contains a solution. To utilize the IVT, we first need to rewrite the equation in the form $f(x) = 0$. This is achieved by subtracting $(x-5)$ from both sides, resulting in the function: $f(x) = \cos(x-5) - (x-5)$. This transformation is crucial because it allows us to search for the roots of the function $f(x)$, which correspond to the solutions of the original equation.

Next, we must establish the continuity of the function $f(x) = \cos(x-5) - (x-5)$. We recognize that $\cos(x-5)$ is a composite function, where the outer function is the cosine function and the inner function is $(x-5)$. The cosine function is continuous for all real numbers, and $(x-5)$ is a linear function, which is also continuous for all real numbers. The composition of continuous functions is continuous, so $\cos(x-5)$ is continuous. Furthermore, $(x-5)$ is a polynomial (a linear function), and polynomials are continuous everywhere. The difference between two continuous functions is also continuous. Therefore, $f(x) = \cos(x-5) - (x-5)$ is continuous for all real numbers. Since $f(x)$ is continuous over all real numbers, it is also continuous on any closed interval.

With the continuity of $f(x)$ confirmed, we can now evaluate $f(x)$ at the endpoints of each given interval. This will allow us to check for sign changes, which, according to the Intermediate Value Theorem, indicate the existence of a root within that interval. Let's begin by analyzing interval A, (3, 4):

• For interval A (3, 4): We evaluate $f(3)$ and $f(4)$.
  • $f(3) = \cos(3-5) - (3-5) = \cos(-2) + 2$. Since cosine is an even function, $\cos(-2) = \cos(2)$. The value of $\cos(2)$ is approximately -0.416. Therefore, $f(3) ≈ -0.416 + 2 ≈ 1.584$, which is positive.
  • $f(4) = \cos(4-5) - (4-5) = \cos(-1) + 1 = \cos(1) + 1$. The value of $\cos(1)$ is approximately 0.540. Therefore, $f(4) ≈ 0.540 + 1 ≈ 1.540$, which is also positive.

Since both $f(3)$ and $f(4)$ are positive, there is no sign change in the interval (3, 4). The IVT does not guarantee a root in this interval. This does not necessarily mean there isn't a root, but the theorem doesn't provide conclusive evidence.

Moving on to interval B (4, 5):

• For interval B (4, 5): We already know $f(4) ≈ 1.540$ (positive). Now we need to calculate $f(5)$.
  • $f(5) = \cos(5-5) - (5-5) = \cos(0) - 0 = 1$. $f(5)$ is also positive.

Again, there is no sign change in the interval (4, 5) as both $f(4)$ and $f(5)$ are positive. Thus, the IVT does not guarantee a root in this interval either.

Let's examine interval C (5, 6):

• For interval C (5, 6): We already know $f(5) = 1$ (positive). Now we evaluate $f(6)$.
  • $f(6) = \cos(6-5) - (6-5) = \cos(1) - 1$. We know that $\cos(1) ≈ 0.540$. Therefore, $f(6) ≈ 0.540 - 1 ≈ -0.460$, which is negative.

Here, we observe a sign change! $f(5)$ is positive, and $f(6)$ is negative. According to the Intermediate Value Theorem, there must be at least one value $c$ in the interval (5, 6) such that $f(c) = 0$. This indicates that there is a solution to the equation $\cos(x-5) = x-5$ within the interval (5, 6).

Finally, we'll check interval D (6, 7) for completeness:

• For interval D (6, 7): We already know $f(6) ≈ -0.460$ (negative). Now we calculate $f(7)$.
  • $f(7) = \cos(7-5) - (7-5) = \cos(2) - 2$. We know that $\cos(2) ≈ -0.416$. Therefore, $f(7) ≈ -0.416 - 2 ≈ -2.416$, which is also negative.

Since both $f(6)$ and $f(7)$ are negative, there is no sign change in the interval (6, 7). The IVT does not guarantee a root in this interval.

Conclusion

Through the application of the Intermediate Value Theorem, we have successfully determined that the equation $\cos(x-5) = x-5$ has a solution within the interval (5, 6). By evaluating the function $f(x) = \cos(x-5) - (x-5)$ at the endpoints of each interval and looking for sign changes, we identified interval C as the one guaranteed to contain a root. This demonstrates the power of the IVT in locating solutions to equations, particularly when analytical methods are difficult to apply. The Intermediate Value Theorem is a powerful tool in calculus, providing a bridge between the theoretical concept of continuity and the practical problem of finding solutions to equations. Its ability to guarantee the existence of solutions within specific intervals makes it an indispensable asset in mathematical analysis and problem-solving.