Problem 8 - Calculus for Biology and Medicine

Question

Let $$ f(x)=\left\\{\begin{array}{cc} \frac{x^{2}+x-2}{x-1} & \text { if } x \neq 1 \\ a & \text { if } x=1 \end{array}\right. $$ Which value must you assign to \(a\) so that \(f(x)\) is continuous at \(x=1 ?\)

Step-by-Step Solution

Answer

The value of \( a \) must be 3 to make \( f(x) \) continuous at \( x = 1 \).

Step 1: Understanding Continuity

A function \( f(x) \) is continuous at \( x = c \) if \( \lim_{x \to c} f(x) = f(c) \). In this exercise, we need to find the value of \( a \) such that \( f(x) \) is continuous at \( x = 1 \). This means we must have \( \lim_{x \to 1} f(x) = f(1) = a \).

Step 2: Find the Limit as x Approaches 1

Since \( f(x) = \frac{x^{2}+x-2}{x-1} \) when \( x eq 1 \), we need to determine \( \lim_{x \to 1} \frac{x^{2}+x-2}{x-1} \). We'll start by simplifying the expression. Note that the numerator \( x^{2} + x - 2 \) can be factored as \( (x-1)(x+2) \).

Step 3: Simplify the Expression

Factor \( x^{2} + x - 2 \) as \( (x-1)(x+2) \). Substitute this factorization into the function:\[\frac{x^{2}+x-2}{x-1} = \frac{(x-1)(x+2)}{x-1}\]For \( x eq 1 \), the \( x-1 \) terms cancel each other out, simplifying the expression to \( x + 2 \).

Step 4: Evaluate the Limit

Now, find the limit:\[\lim_{x \to 1} (x + 2) = 1 + 2 = 3\]Thus, \( \lim_{x \to 1} f(x) = 3 \).

Step 5: Determine the Value of a

For the function \( f(x) \) to be continuous at \( x = 1 \), we need \( f(1) = a = 3 \), matching the limit we found in step 4. Therefore, the value of \( a \) that ensures continuity is \( 3 \).

Key Concepts

Limits in Calculus

Limits in calculus is one of the foundational concepts. Think of it as understanding what happens to a function as the inputs get closer and closer to some number. In more technical terms, the limit \( \lim_{x \to c} f(x) \) is the value that \( f(x) \) approaches as \( x \) approaches \( c \.\)
- Limits help determine continuity of a function.- They analyze the computational behavior of expressions and functions, especially where direct evaluation is difficult.In our exercise, to ensure continuity at \( x = 1 \,\) we determined the limit of \( f(x) \) as \( x \to 1 \) and found it to be 3. This suggests addressing what value \( a \) must take when the function \( f(x) \) becomes a single value, as \( x \) directly approaches 1.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. These functions often have vertical asymptotes and holes, where the denominator equals zero.
For example, in the function \( f(x) = \frac{x^2 + x - 2}{x - 1} \,\) the denominator becomes zero at \( x = 1 \,,\) indicating a potential problem at \( x = 1 \.\) However, if we can cancel out the factor causing the zero in the denominator, it becomes easier to analyze the function behavior.- When a rational function has common factors in the numerator and denominator, we can often simplify it.- Simplifying can help to detect removable discontinuities, also known as 'holes.'
In our exercise, factorization avoided the discontinuity by canceling out the \( x-1 \) term, simplifying the function.

Factorization

Factorization is a technique used to simplify algebraic expressions. By breaking down a polynomial into products of simpler polynomials, it becomes easier to manipulate and understand these expressions. For the exercise at hand, factorization played a crucial role in simplifying the rational function.
- Factorization transforms \( x^2 + x - 2 \,\) a quadratic expression, into \( (x-1)(x+2) \.\)- This allows for the identification of common factors in the numerator and denominator.- By cancelling out these common factors, one can remove removable discontinuities in rational functions.
Using factorization in this case allowed us to cancel the \( x-1 \) in the numerator and denominator, simplifying \( f(x) \,\) thus letting us determine the continuity at \( x = 1 \.\) This shows how powerful factorization is in handling limits and continuity issues.