Problem 5 - Calculus for Biology and Medicine

Question

Use the Intermediate Value Theorem to show that \(e^{-x}=x^{2}\) has a solution in \((0,1)\).

Step-by-Step Solution

Answer

By the Intermediate Value Theorem, since \( f(0) > 0 \) and \( f(1) < 0 \), there is a solution in \((0, 1)\).

Step 1: Define the Function

Let's define the function that we will analyze using the Intermediate Value Theorem. Let \( f(x) = e^{-x} - x^2 \). We want to find if there is some \( c \in (0, 1) \) such that \( f(c) = 0 \).

Step 2: Evaluate the Function at the Endpoints

Compute the values of \( f(x) \) at the endpoints of the interval \[ f(0) = e^{0} - 0^2 = 1 \]\[ f(1) = e^{-1} - 1^2 = \frac{1}{e} - 1 \]Since \( \frac{1}{e} \approx 0.3679 \), we find that \( f(1) \approx 0.3679 - 1 = -0.6321 \). Thus, \( f(0) = 1 \) and \( f(1) < 0 \).

Step 3: Apply the Intermediate Value Theorem

The Intermediate Value Theorem states that if a function \( f \) is continuous on a closed interval \([a, b]\), and \( f(a) \) and \( f(b) \) have opposite signs, then there exists at least one \( c \in (a, b) \) such that \( f(c) = 0 \). Here, \( f(x) = e^{-x} - x^2 \) is continuous in \([0, 1]\) since both \( e^{-x} \) and \( x^2 \) are continuous functions. Since \( f(0) > 0 \) and \( f(1) < 0 \), there must be some \( c \in (0, 1) \) such that \( f(c) = 0 \).

Step 4: Conclusion

Since we have shown that \( f(x) \) is continuous and the signs of \( f(0) \) and \( f(1) \) are opposite, by the Intermediate Value Theorem, there exists a solution to \( e^{-x} = x^2 \) within the interval \((0, 1)\).

Key Concepts

Continuous Functions

In mathematics, a function is considered continuous if it does not have any abrupt changes, breaks, or holes in its graph. This means that if you were to draw the graph of the function, you could do it without lifting your pencil from the paper.
For the function defined in our exercise, \( f(x) = e^{-x} - x^2 \), we analyze it on the interval \([0, 1]\). Both \( e^{-x} \) and \( x^2 \) are separately continuous functions.
- The function \( e^{-x} \) is an exponential function, which is smooth and continuous over all real numbers.- On the other hand, \( x^2 \) is a simple polynomial, making it also continuous everywhere.

• When combining these two functions through addition or subtraction, the result \( f(x) = e^{-x} - x^2 \) maintains its continuity.

Recognizing a function's continuity is essential, especially when employing the Intermediate Value Theorem, as it requires the function to be continuous on the considered interval.

Root Finding

Finding the root of a function means finding the value of \( x \) at which the function equals zero, i.e., \( f(x) = 0 \). When we can't solve a function analytically, like in our case of \( f(x) = e^{-x} - x^2 \), numerical methods or theorems, such as the Intermediate Value Theorem (IVT), can help us identify roots.
For the given problem, the root finding process involves checking the sign of the function at the endpoints of a specified interval:

• First, \( f(0) \) is evaluated and found to be positive, \( f(0) = 1 \).
• Next, \( f(1) \) is evaluated and found to be negative, \( f(1) \approx -0.6321 \).

The change in sign from positive to negative across the interval \([0, 1]\) suggests at least one root in the open interval \((0, 1)\), thanks to the Intermediate Value Theorem.

Function Evaluation

Evaluating a function involves calculating its value at specific points to understand its behavior. In applying the Intermediate Value Theorem, examining the function at the endpoints of the interval is crucial.
For \( f(x) = e^{-x} - x^2 \), you evaluate:

• At \( x = 0 \): \( f(0) = e^0 - 0^2 = 1 \).
• At \( x = 1 \): \( f(1) = e^{-1} - 1^2 \), which is approximately \( -0.6321 \).

This evaluation process provides critical information about the behavior of the function over the interval \([0, 1]\). Seeing that \( f(0) \) and \( f(1) \) have opposite signs hints at the root's existence due to their continuity. Function evaluation is thus a fundamental tool in applying the Intermediate Value Theorem effectively, especially in root finding problems.