Problem 5 - Calculus for Biology and Medicine

Question

(a) Show that, for \(x \neq 1\), $$ \frac{x^{2}-1}{x+1}=x-1 $$ (b) Are the functions \(f(x)\) and \(g(x)\) equal? $$ \begin{array}{l} f(x)=\frac{x^{2}-1}{x+1}, \quad x \neq-1 \\ g(x)=x-1, \quad x \in \mathbf{R} \end{array} $$

Step-by-Step Solution

Answer

(a) Yes, \( \frac{x^2 - 1}{x+1} = x-1 \) for \(x \neq 1\). (b) No, the functions are not equal due to different domains.

Step 1: Simplify the Expression in (a)

The expression \( \frac{x^2 - 1}{x+1} \) can be simplified. Notice that the numerator \(x^2 - 1\) can be factored as \((x-1)(x+1)\) because it is a difference of squares. Therefore, the expression becomes: \[ \frac{(x-1)(x+1)}{x+1} \] Now, we can cancel \(x+1\) from the numerator and denominator: \[ \frac{(x-1)\cancel{(x+1)}}{\cancel{(x+1)}} = x-1 \] Thus, for \(x eq 1\), \( \frac{x^2 - 1}{x+1} = x - 1 \).

Step 2: Verify if Functions are Equal

To determine if functions \(f(x)\) and \(g(x)\) are equal, we need to consider their domains. Function \( f(x) = \frac{x^2 - 1}{x+1} \) is defined for \(x eq -1\), meaning it is undefined at \(x = -1\). Function \( g(x) = x - 1 \) is defined for all \(x \in \mathbf{R}\). Even though the expressions are algebraically equivalent, their domains differ. Therefore, \(f(x)\) and \(g(x)\) are not equal as functions, due to the difference in their domains.

Key Concepts

Difference of Squares

A difference of squares refers to an algebraic expression of the form \(a^2 - b^2\), which can be factored into \((a-b)(a+b)\). This is a useful identity that can simplify expressions significantly.
In the exercise, the expression \(x^2 - 1\) is a difference of squares. Here, \(a=x\) and \(b=1\), so it factors to \((x-1)(x+1)\). This technique allows us to simplify complex expressions and is especially helpful in calculus for evaluating limits, derivatives, and integrals involving polynomial terms.

• Recognizing a difference of squares makes simplification straightforward.
• Factoring polynomials where applicable can uncover simpler relationships between functions.

Mastering this technique will aid in handling more complex calculus problems efficiently.

Function Equality

Function equality asks whether two functions express the same rule for assigning outputs to inputs, while also having the same domain. Even if two expressions simplify to the same form, they are not considered equal unless their domains match.
In the exercise, while \(f(x) = \frac{x^2 - 1}{x+1}\) simplifies to \(x-1\), the functions \(f(x)\) and \(g(x) = x-1\) are not equal because their domains differ. Function equality requires both identical outputs for all inputs and the same set of permissible inputs.

• Always check the domain first when considering function equality.
• Even algebraically identical expressions can represent different functions due to domain restrictions.

Understanding this concept is crucial in calculus to correctly interpret the behavior of functions across their entire use.

Domain of Functions

The domain of a function is the set of input values (\(x\)-values) for which the function is defined. It is essential to identify which \(x\)-values cause the function to be undefined, such as division by zero or the square root of a negative number.
In the given problem, \(f(x) = \frac{x^2 - 1}{x+1}\) is undefined when \(x = -1\), since it causes division by zero. Therefore, its domain excludes \(x = -1\). Conversely, \(g(x) = x-1\) has a domain of all real numbers, \(\mathbb{R}\).

• Always examine the denominator for potential zero-values.
• Analyze other expressions (such as square roots) for potential domain restrictions.

Knowing how to determine the domain is fundamental for analyzing where any function can be applied safely.