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 $$ f(x)=\frac{x^{2}-1}{x-1}, \quad x \neq 1 $$ and $$ g(x)=x+1, \quad x \in \mathbf{R} $$ equal?

Step-by-Step Solution

Answer

Functions are equal for all \( x \neq 1 \); \( f(x) \) is undefined at \( x = 1 \).

Step 1: Understand the Problem

We need to prove that the expression \( \frac{x^{2}-1}{x-1} \) can be simplified to \( x+1 \) when \( x eq 1 \). Then, we should check if the functions \( f(x)=\frac{x^{2}-1}{x-1} \) for \( x eq 1 \) and \( g(x)=x+1 \) for all real \( x \) are equal.

Step 2: Simplify the Arithmetic Expression

For (a), recognize the numerator \( x^2 - 1 \) as a difference of squares. Use the factorization formula, where \( a^2 - b^2 = (a-b)(a+b) \). So, \( x^2 - 1 = (x - 1)(x + 1) \).

Step 3: Cancel Common Factors

For \( f(x) = \frac{x^{2}-1}{x-1} \), after writing \( x^{2} - 1 \) as \((x - 1)(x + 1)\), it becomes \( \frac{(x - 1)(x + 1)}{x - 1} \). Since \( x eq 1 \), cancel the common factor \( x - 1 \), yielding \( f(x) = x + 1 \).

Step 4: Compare Functions

In (b), compare \( f(x) = x + 1 \) and \( g(x) = x + 1 \) for values of \( x eq 1 \). Since both expressions simplify to \( x + 1 \) under these conditions, they are equal for all \( x \) except \( x = 1 \), where \( f(x) \) is undefined.

Key Concepts

Functions

Functions are one of the fundamental building blocks in calculus. A function, at its core, is a relation between a set of inputs and a set of possible outputs, where each input is related to exactly one output. For any function, we can say it maps or transforms an input, often denoted by \( x \), to an output value. In the context of the original exercise, we are comparing two functions: \( f(x) \) and \( g(x) \).

• \( f(x) = \frac{x^{2}-1}{x-1} \) for \( x eq 1 \)
• \( g(x) = x + 1 \) for all real \( x \)

Both functions generate an output based on the given \( x \). Understanding functions helps us in analyzing how these outputs change as the inputs change, and in finding if two different-looking functions actually try to tell us the same thing about their outputs.

Simplification

Simplification in calculus often means reducing expressions into a more manageable or understandable form. It doesn't change the essence or value but makes it easier to work with. In the given exercise, simplifying \( \frac{x^2 - 1}{x - 1} \) to \( x + 1 \) involves recognizing patterns and using algebraic rules. Here:- We use the difference of squares, a special algebraic identity, to factorize the expression.- Once factored, we can cancel out terms to simplify the expression.Simplification helps to ensure that we can compare functions efficiently as it boils down expressions to their simplest form, enabling easier computation and understanding of what's happening within equations.

Factorization

Factorization is a method used to break down expressions into products of simpler expressions or factors. It's a powerful tool in algebra used to simplify expressions or solve equations. In our exercise, the expression \( x^2 - 1 \) can be factorized using the identity for the difference of squares:\[ x^2 - 1 = (x - 1)(x + 1) \]

• This identity states that any expression of the form \( a^2 - b^2 \) can be broken down into the product \( (a-b)(a+b) \).
• By applying this, we turn a complex quadratic expression into simpler linear factors.

Factorization thus allows us to handle expressions more easily, especially when simplifying fractions or solving equations.

Difference of Squares

The difference of squares is a specific pattern in algebra that simplifies quickly into the product of two binomials. Recognizing and using this identity is crucial in simplification and factorization tasks. The formula is:\[ a^2 - b^2 = (a - b)(a + b) \]For our exercise, this means:- Identify \( a \) as \( x \) and \( b \) as 1.- Thus, the expression \( x^2 - 1 \) fits this pattern and can be rewritten as \( (x - 1)(x + 1) \).Knowing and applying the difference of squares helps transform complex algebraic expressions to simpler forms, which is pivotal in calculus for simplifying functions and expressions effectively.