Problem 53 - Calculus: Early Transcendentals

Question

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. \( x^4 + x - 3 = 0 \), \( (1, 2) \)

Step-by-Step Solution

Answer

There is a root in the interval \((1, 2)\) by the Intermediate Value Theorem.

Step 1: Understanding the Intermediate Value Theorem

The Intermediate Value Theorem states that if a function \( f \) is continuous on the interval \([a, b]\) and \( N \) is a number between \( f(a) \) and \( f(b) \), then there exists at least one \( c \) in the interval \((a, b)\) such that \( f(c) = N \). In simpler terms, if a function is continuous on a certain interval and changes from one value to another, it must cross every value in between at least once.

Step 2: Define the Function

Let \( f(x) = x^4 + x - 3 \). We need to show that there is at least one root within the interval \((1, 2)\). This means finding some \( c \) in \((1, 2)\) where \( f(c) = 0 \).

Step 3: Evaluate the Function at the Endpoints

Calculate \( f(1) \) and \( f(2) \). - \( f(1) = 1^4 + 1 - 3 = 1 + 1 - 3 = -1 \)- \( f(2) = 2^4 + 2 - 3 = 16 + 2 - 3 = 15 \)

Step 4: Check for a Sign Change

Since \( f(1) = -1 \) and \( f(2) = 15 \), and the values of \( f \) at these points are of opposite signs, by the Intermediate Value Theorem, there must be some value \( c \) in the interval \((1, 2)\) such that \( f(c) = 0 \).

Step 5: Conclude the Existence of a Root

With a sign change from negative to positive between \( x = 1 \) and \( x = 2 \), and knowing \( f(x) \) is continuous (since it's a polynomial), there must exist at least one root in the interval \((1, 2)\). Therefore, the Intermediate Value Theorem confirms this root.

Key Concepts

Continuous Function

A continuous function is one that does not "break" or "jump" on an interval. Imagine drawing a curve with no interruptions between two points. That's what continuity is about. For a function to be continuous on an interval, it must have no gaps, filled holes, or jumps.
In the context of the Intermediate Value Theorem, continuity is crucial. If a function is continuous over an interval, like our polynomial function, it means we can trust the function to follow predictable behavior without any surprises.
Whenever a polynomial function is involved, rest assured that it's a continuous function across its domain, making it a perfect candidate for applying the Intermediate Value Theorem.

Root of Equation

In mathematical terms, a "root" of an equation is a value that, when substituted into the equation, results in 0. Essentially, it's where the graph of the function crosses the x-axis.
For the equation given, \(x^4 + x - 3 = 0\), the root is the value \(c\) within the specified interval, \((1, 2)\), that satisfies \(f(c) = 0\).
Finding a root is like solving a puzzle to pin down where the function value transitions through zero. The Intermediate Value Theorem helps locate such roots by ensuring a crossing point in the graph.

Polynomial Function

Polynomial functions are expressions consisting of variables and coefficients, combined through addition, subtraction, multiplication, but not division by variables. They are written as \(f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\).
One important property of polynomial functions is that they are continuous everywhere. This is a key reason why the Intermediate Value Theorem applies so well to them.
For the example in hand, \(x^4 + x - 3\) is a polynomial function, and thus we can guarantee that it maintains continuity between any two points on the x-axis.

Sign Change in Interval

The concept of a sign change in an interval is central to applying the Intermediate Value Theorem. If a function changes sign between two points on an interval, it indicates passing through zero, suggesting a root.
In our example, by calculating \(f(1) = -1\) and \(f(2) = 15\), we observe the function value goes from negative at \(x = 1\) to positive at \(x = 2\).
This sign change flags that somewhere in the interval \((1, 2)\), the function value must cross zero. The Intermediate Value Theorem then ensures there must be at least one root between 1 and 2, confirming a crossing point.