Problem 60 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the Intermediate Value Theorem to show that \(x^{3}-7 x^{2}+14 x-8=0\) has at least one solution in the interval \([0,5] .\) Sketch the graph of \(y=x^{3}-7 x^{2}+14 x-8\) over \([0,5]\). How many solutions does this pauation really have?

Step-by-Step Solution

Answer

The function has at least one root in \([0,5]\), and actually three solutions.

Step 1: Understand 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 the interval \((a, b)\) such that \( f(c) = 0 \). This means the function crosses the x-axis somewhere in the interval.

Step 2: Define the Function

The given function is \(f(x) = x^{3} - 7x^{2} + 14x - 8\). We need to show it has at least one root in the interval \([0, 5]\). The function involves polynomial operations, and since polynomials are continuous everywhere, \(f\) is continuous on \([0, 5]\).

Step 3: Calculate Values at Interval Endpoints

Evaluate the function at the endpoints of the interval. First, calculate \(f(0)\) and \(f(5)\):\[f(0) = 0^3 - 7 \cdot 0^2 + 14 \cdot 0 - 8 = -8\] \[f(5) = 5^3 - 7 \cdot 5^2 + 14 \cdot 5 - 8 = 125 - 175 + 70 - 8 = 12\].

Step 4: Verify Opposite Signs

Check the signs of \(f(0)\) and \(f(5)\). We have \(f(0) = -8\) (negative) and \(f(5) = 12\) (positive). Since \(f(0)\) and \(f(5)\) have opposite signs, by the Intermediate Value Theorem, there must be at least one \(c\) in \((0, 5)\) with \(f(c) = 0\).

Step 5: Sketch the Graph

To provide a visual understanding, sketch the graph of \(y = f(x) = x^3 - 7x^2 + 14x - 8\) over the interval \([0, 5]\). You should see the curve starting below the x-axis at \(x = 0\) and crossing above the x-axis indicating at least one root.

Step 6: Additional Roots

To find the actual number of roots, observe the graph more carefully or use a software tool. From calculations or a detailed graphing, it is determined that the function actually crosses the x-axis 3 times in \([0, 5]\), indicating three real solutions.

Key Concepts

Polynomials

Polynomials are expressions involving variables and coefficients, combined using operations of addition, subtraction, multiplication, and non-negative integer exponents. A basic form of a polynomial is \(a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0\), where \(a_n, a_{n-1}, \ldots, a_0\) are constants, and each \(x\) is raised to a power that is a whole number. Since they are built upon simple arithmetic operations and powers, they exhibit smooth curves when graphed.

• Appear as smooth, continuous graphs without breaks or gaps.
• Can represent various degrees, like linear (degree 1), quadratic (degree 2), cubic (degree 3), and so forth.

For instance, the function given in the exercise, \(f(x) = x^3 - 7x^2 + 14x - 8\), is a cubic polynomial. This means it is characterized by a degree of 3, reflected in its highest power of "x", which results in its ability to change direction up to twice, creating a maximum of two bends in its graph as it moves on a Cartesian plane.

Continuous Functions

A continuous function is one that does not have unexpected jumps or breaks in its behavior. You can think of a function as continuous if you can draw it on a graph without lifting your pencil. Polynomials, like the one in the exercise, are classically continuous functions.

• They do not have any breaks, jumps, or holes in their graph.
• They allow the application of the Intermediate Value Theorem, making analysis straightforward.

The Intermediate Value Theorem, a powerful tool in calculus, applies to continuous functions. It postulates that for any value \(y\) between \(f(a)\) and \(f(b)\) on an interval \([a, b]\), there exists at least one point \(c\) in \((a, b)\) where \(f(c) = y\). In the problem, we used this theorem to prove the existence of a solution within the interval [0, 5]. We confirmed that the polynomial is continuous, validating the usage of the theorem.

Roots of Equations

The root of an equation is a value of \(x\) that makes the function equal to zero. Finding roots, or solutions, is often a primary objective in mathematical problems involving polynomials. These roots correspond to the x-coordinates where the graph of a function intersects the x-axis.

• Helps determine where polynomials reach zero, important for applications in physics, engineering, and other sciences.
• Can have multiple real roots, even in a single interval.

In the exercise, we identified an interval \([0, 5]\) where the polynomial \(x^3 - 7x^2 + 14x - 8\) changes sign from negative to positive as proven by the Intermediate Value Theorem. This change of sign confirms at least one root within this interval. By graphing the polynomial or using numerical methods, it's revealed that there are actually three roots in this interval, depicting multiple instances where the function intersects the x-axis.