Problem 55

Question

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. $ e^x = 3 - 2x $, $ (0, 1) $

Step-by-Step Solution

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Answer

There is a root in (0, 1) because the function changes signs there.

1Step 1: Understand the problem

We need to show that there is a root for the equation $ e^x = 3 - 2x $ within the interval (0, 1) using the Intermediate Value Theorem (IVT). This involves defining a continuous function that changes signs over this interval.

2Step 2: Define a function

Create a function $ f(x) = e^x - (3 - 2x) $. A root of $ e^x = 3 - 2x $ is equivalent to finding $ x $ such that $ f(x) = 0 $.

3Step 3: Verify continuity

The function $ f(x) $ is made up of $ e^x $, which is continuous, and the polynomial $-3 + 2x$, which is also continuous. Thus, $ f(x) $ is continuous across all $ x $, including the specified interval $(0, 1)$.

4Step 4: Evaluate endpoints of the interval

Calculate $ f(0) $ and $ f(1) $.\[ f(0) = e^0 - (3 - 0) = 1 - 3 = -2 \]\[ f(1) = e^1 - (3 - 2) = e - 1 \approx 2.718 - 1 = 1.718 \]

5Step 5: Apply the Intermediate Value Theorem

Since $ f(0) = -2 $ and $ f(1) = 1.718 $, $ f(x) $ changes signs from negative to positive over the interval $(0, 1)$. By the IVT, because $ f(x) $ is continuous, there exists at least one root in the interval where $ f(x) = 0 $.

Key Concepts

Continuous FunctionsRoots of EquationsFunction Evaluation

Continuous Functions

Continuous functions are at the heart of the Intermediate Value Theorem (IVT). When mathematicians say a function is continuous, they mean there are no breaks, jumps, or holes in its graph. It can be drawn without lifting your pen from the paper. This property of smoothness is crucial for applying the IVT. For the function in our exercise,

The exponential part, denoted as $ e^x $, is a classic example of a continuous function. It rises smoothly and exponentially keeps moving upwards as $ x $ increases.
The polynomial part, $-3 + 2x$, is likewise continuous; it’s simply a straight line with a steady slope.

Together, they form the function $ f(x) = e^x - (3 - 2x) $. This combined function retains continuity over every point, crucially over the interval (0, 1) for our purposes in solving the problem. By understanding these properties, we can confidently apply the IVT knowing that our function behaves predictably and smoothly in this range.

Roots of Equations

The idea of finding the root of an equation is simply about determining the value that makes the equation true, or in simpler terms, makes it equal zero. For the equation $ e^x = 3 - 2x $, a root occurs when two expressions equal each other. By transforming the equation into a single function $ f(x) = e^x - (3 - 2x) $, finding its root translates to finding where $ f(x) = 0 $.This is where the Intermediate Value Theorem helps us. Consider:

We calculated $ f(0) = -2 $, which is negative.
We calculated $ f(1) = 1.718 $, which is positive.

The situation where a negative value transitions to a positive indicates that there must be at least one point in between where $ f(x) = 0 $—this is our root. It means there exists a specific $ x $ within the interval (0, 1) that solves our original equation.

Function Evaluation

Function evaluation is straightforward but essential when working with equations and their roots. It involves substituting specific values into a function to find out what it yields. In the context of using the Intermediate Value Theorem, evaluating the endpoints of our interval was a key step.Here’s how it helps:

At $ x = 0 $, replacing $ x $ in $ f(x) = e^x - (3 - 2x) $ gives $ f(0) = 1 - 3 = -2 $.
At $ x = 1 $, it simplifies to $ f(1) = e - 1 \approx 1.718 $.

By determining these specific values, we demonstrate that $ f(x) $ changes from negative to positive across the interval. This change confirms that the Intermediate Value Theorem is applicable, proving the existence of a zero in the function, indicating the presence of a root within the interval. Function evaluation thus offers a powerful confirmation technique for understanding the behavior of equations in real-world scenarios.

Problem 55

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