Problem 11 - Calculus

Question

show that there is a number \(c,\) with \(0 \leq c \leq 1,\) such that \(f(c)=0\) $$f(x)=x^{3}+x^{2}-1$$

Step-by-Step Solution

Answer

There exists a number \\(c \\in [0, 1]\\) such that \\(f(c) = 0\\).

Step 1: Evaluate f at the endpoints

\(f(0) = 0 + 0 - 1 = -1 < 0\)
\(f(1) = 1 + 1 - 1 = 1 > 0\)

Step 2: Apply the Intermediate Value Theorem

Since \(f\) is a polynomial, it is continuous on \([0, 1]\). Since \(f(0) = -1 < 0 < 1 = f(1)\), by the Intermediate Value Theorem, there exists \(c \in (0, 1)\) such that \(f(c) = 0\). \(\square\)

Key Concepts

root-finding

Root-finding is an essential concept in mathematics. It involves discovering the values of a variable that make a function equal zero. This is especially relevant when analyzing functions like polynomial equations. Finding these roots helps in understanding the behavior and characteristics of the polynomial itself.

Methods like the Intermediate Value Theorem can assist in root-finding by ensuring that if a change in sign is detected on a continuous interval, a root exists within that interval. Hence, for the function given in the exercise, acknowledging this theorem helps us to locate a root between 0 and 1. This root is where the function value changes from negative to positive, providing a solution.

polynomial function

A polynomial function is a mathematical expression involving sums of powers in one or more variables. The standard form is generally written as: \( f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 \), where the coefficients \( a_i \) are real numbers, and \( i \) is a non-negative integer.

In our exercise, the function is \( f(x) = x^3 + x^2 - 1 \). This is a cubic polynomial because the highest power of \( x \) is three. Polynomial functions are important as they can model a variety of real-world situations, such as the trajectory of a projectile or predicting income over time. By understanding polynomial functions, solving problems involving real-world applications becomes easier.

sign change

Sign change in a mathematical function refers to a transition from positive to negative values or vice versa as the input variable changes. This concept is crucial when looking for roots of a polynomial function because it indicates the presence of a solution or root in the interval where the change occurs.

In the exercise example, evaluating \( f(0) = -1 \) and \( f(1) = 1 \) confirms a sign change between \( x = 0 \) and \( x = 1 \). This indicates there must be a root somewhere in the interval, according to the Intermediate Value Theorem, thus confirming the presence of number \( c \) such that \( f(c) = 0 \). The detection of this sign change is facilitated by evaluating the function at interval endpoints.

continuous function

A continuous function is a function without any breaks, gaps, or jumps in its graph. Mathematical continuity ensures if there is a change in a function's output, nearing a given point, there will be an associated change in its input.

The Intermediate Value Theorem, used in the solution for this exercise, relies heavily on continuity. The theorem states that for any continuous function \( f \) over an interval \([a, b]\), if \( f(a) \) and \( f(b) \) have opposite signs, then there exists a point \( c \) within that interval where \( f(c) = 0 \).

In the polynomial \( f(x) = x^3 + x^2 - 1 \), being a polynomial confirms continuity across its domain. This property is what legitimizes the theorem in determining the existence of a point \( c \) where the function's value is zero. Understanding continuous functions allows us to apply this theorem effectively in finding roots.