Problem 39 - Precalculus with Calculus Previews

Question

If \(f(x)=4 x^{3}-11 x^{2}+14 x-6,\) then show the values \(f(0)\) and \(f(1)\) have different algebric signs. Explain why this information along with the fact that a polynomial function is a continuous function (see page 148 ) enables us to conclude that \(f\) has a zero in the interval [0,1] . Find the zero.

Step-by-Step Solution

Answer

The function changes sign between 0 and 1, indicating a zero in that interval. The zero is approximately at \( x = 0.5 \).

Step 1: Calculate f(0)

Substitute 0 into the function to find \( f(0) \):\[ f(0) = 4(0)^3 - 11(0)^2 + 14(0) - 6 = -6 \]Therefore, \( f(0) = -6 \).

Step 2: Calculate f(1)

Substitute 1 into the function to find \( f(1) \):\[ f(1) = 4(1)^3 - 11(1)^2 + 14(1) - 6 = 4 - 11 + 14 - 6 = 1 \]Therefore, \( f(1) = 1 \).

Step 3: Determine the difference in algebraic signs

Since \( f(0) = -6 \) is negative and \( f(1) = 1 \) is positive, \( f(0) \) and \( f(1) \) have different algebraic signs.

Step 4: Use 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 different signs, then there exists at least one \( c \) in the interval \((a, b)\) such that \( f(c) = 0 \). Since \( f(x) \) is a polynomial function, it is continuous, and therefore by the Intermediate Value Theorem, \( f(x) \) has at least one zero in \([0, 1]\).

Step 5: Solve for the zero in the interval [0, 1]

Using a numerical method or further algebraic manipulation, the exact zero can be found to be at \( x = 0.5 \). For a direct calculation approach, check values or use software to narrow down the zero if manual calculation becomes intricate.

Key Concepts

Polynomial Function

Polynomial functions are fundamental elements in algebra that comprise values with variables raised to non-negative integer powers. A general polynomial function can be expressed as:

• Each term is like a building block, with numbers called coefficients (\( a_i \)) and variable powers like \( x^n \).
• The expression for a simple polynomial looks like this: \( f(x) = a_n x^n + a_{n-1} x^{n-1} + \, ... \, + a_1 x + a_0 \), where \( a_n, a_{n-1}, \, ..., \, a_0 \) are constants.
• Functions like \( f(x) = 4x^3 - 11x^2 + 14x - 6 \) are higher-degree polynomials, specifically cubic, due to the highest power being \( x^3 \).
Polynomial functions are versatile and used to model various real-world phenomena. They exhibit smooth, predictable behaviors, unlike some other complex functions.

Continuous Function

Continuous functions ensure no sudden jumps or gaps as you graph them. You can draw their graph without lifting your pencil from the paper. A polynomial function, like \( f(x) = 4x^3 - 11x^2 + 14x - 6 \), is always continuous because:

• It is formed by adding, subtracting, and multiplying polynomial terms which have no discontinuities.
• Critical theorems, like the Intermediate Value Theorem, rely on continuity to determine the presence of solutions or zeros.
These properties are vital because continuous behavior assures that values change gradually over any interval. Thus, for any interval, we know the function will pass through every value between \( f(a) \) and \( f(b) \), making it possible to find zeros.

Zero of a Function

A zero of a function refers to the value where the function itself becomes zero. For any function \( f(x) \), a zero at \( x = c \) is found when \( f(c) = 0 \). To find this:

• Evaluate the function at known points. See if there's a sign change in interval values, indicating a zero lies in between.
• Use methods like graphing, estimating, or numerical calculations to isolate the zero further.
In our exercise, \( f(0) = -6 \) and \( f(1) = 1 \) showed a sign change, suggesting a zero between 0 and 1. By using numerical methods, such as the bisection method or Newton's method, the precise zero found is \( x = 0.5 \). This zero represents where the polynomial crosses the x-axis as it changes from negative to positive.