Problem 77 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

In Exercises 77-80, (a) show that the function \(f\) is continuous for all values of \(x\) in the interval \([a, b]\) and (b) prove that \(f\) must have at least one zero in the interval \((a, b)\) by showing that \(f(a)\) and \(f(b)\) have opposite signs. \(f(x)=x^{2}-6 x+8 ; a=1, b=3\)

Step-by-Step Solution

Answer

The function \(f(x) = x^2 - 6x + 8\) is continuous in the interval \([1, 3]\) as it is a polynomial function. By evaluating \(f(1) = 3 > 0\) and \(f(3) = -1 < 0\), we see that \(f(a)\) and \(f(b)\) have opposite signs. Therefore, by applying the Intermediate Value Theorem, there must be at least one zero in the interval \((1, 3)\).

Step 1: Show continuity of the function in the interval [a, b]

A polynomial function is always continuous on its entire domain, which includes all Real numbers. Therefore, the function f(x) = x^2 - 6x + 8 is continuous in the interval [1, 3].

Step 2: Calculate f(a) and f(b)

Calculate the values of the function at a = 1 and b = 3. f(a) = f(1) = (1)^2 - 6(1) + 8 f(b) = f(3) = (3)^2 - 6(3) + 8

Step 3: Evaluate the signs of f(a) and f(b)

Evaluate and check the signs of f(a) and f(b): f(1) = 1 - 6 + 8 = 3 f(3) = 9 - 18 + 8 = -1 Since f(a) = 3 > 0 and f(b) = -1 < 0, f(a) and f(b) have opposite signs.

Step 4: Apply the Intermediate Value Theorem

Since the function f(x) = x^2 - 6x + 8 is continuous in the interval [a, b] and f(a) and f(b) have opposite signs, we can now apply the Intermediate Value Theorem. According to the theorem, there must exist at least one point c in the interval (a, b) such that f(c) = 0. This means that the function f(x) must have at least one zero in the interval (1, 3).

Key Concepts

Polynomial Functions

Polynomial functions form a broad class of mathematical expressions involving variables with non-negative integer powers. These functions, like the one in our example exercise, are expressed in the form \( a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 \), where each \(a\) represents a coefficient and \(n\) is a non-negative integer.

• For our function \(f(x) = x^2 - 6x + 8\), it's a quadratic polynomial because it is of degree 2 (the highest power of \(x\) is 2).
• Polynomial functions are incredibly versatile and include cases like linear, quadratic, cubic, and higher-degree polynomials.

Polynomial functions are continuous everywhere, which means they have no breaks, jumps, or gaps. This property is crucial for the concepts discussed further. Since these functions span across all real numbers, they are defined and smooth throughout their entire expression.

Continuity

Continuity is a fundamental property in calculus that ensures a function behaves predictably over an interval. A function is continuous if it can be graphed without lifting your pencil off the paper. For our polynomial function \(f(x) = x^2 - 6x + 8\), this idea holds true for all real numbers, including the interval \([1, 3]\).

• Being continuous on \([1, 3]\) implies that for any small change in \(x\), the change in \(f(x)\) is equally small, indicating no abrupt jumps or breaks.
• This seamless flow makes polynomial functions reliable for analysis, especially when applying theorems like the Intermediate Value Theorem.

In terms of practical usage, continuity allows for predictions and computations of intermediate values accurately, without missing key points within the specified range of \(x\).

Zero of a Function

The zero of a function is a valuable concept, representing the \(x\)-value where the output of the function, \(f(x)\), equals zero. This is sometimes called the root of the function. In our example function, \(f(x) = x^2 - 6x + 8\), our task is to identify a zero within the interval \((1, 3)\) using the Intermediate Value Theorem.

• To start, we evaluate \(f(1)\) and \(f(3)\). Here, \(f(1) = 3\) and \(f(3) = -1\), showing opposite signs, one positive and one negative.
• This sign change indicates that between these two points, the function must cross the x-axis at least once.

According to the Intermediate Value Theorem, if a continuous function like ours changes from positive to negative, or vice versa, there is at least one \(c\) in the interval \((1, 3)\) such that \(f(c) = 0\). This theorem provides a guaranteed method to confirm that a zero exists within 1 and 3, affirming the concept of continuity and how it aids in determining function behavior.