Question

Determine if \(f\) is continuous at the indicated values. If not, explain why. \(f(x)=\left\\{\begin{array}{cl}\frac{x^{2}+5 x+4}{x^{2}+3 x+2} & x \neq-1 \\\ 3 & x=-1\end{array}\right.\) (a) \(x=-1\) (b) \(x=10\)

Short answer

The function is continuous at both \(x = -1\) and \(x = 10\).

Step-by-step solution

Define Continuity at a Point

A function is continuous at a point \(x = c\) if three conditions are satisfied: the function \(f(x)\) is defined at \(c\), the limit \(\lim_{x \to c} f(x)\) exists, and \(\lim_{x \to c} f(x) = f(c)\). We will examine these conditions for both \(x = -1\) and \(x = 10\).

Check Continuity at \(x = -1\)

**1. Function Defined at \(x = -1\):**\The piecewise definition gives \(f(-1) = 3\), so it is defined.\**2. Calculate the Limit as \(x \to -1\):**\The function \(f(x) = \frac{x^2 + 5x + 4}{x^2 + 3x + 2}\) when \(x eq -1\). Factor both numerator and denominator: \[x^2 + 5x + 4 = (x + 1)(x + 4)\] \[x^2 + 3x + 2 = (x + 1)(x + 2)\].\Cancel \((x + 1)\) to simplify: \(\lim_{x \to -1} \frac{(x + 1)(x + 4)}{(x + 1)(x + 2)} = \lim_{x \to -1} \frac{x + 4}{x + 2}\). Substitute \(x = -1\): \(\lim_{x \to -1} \frac{-1 + 4}{-1 + 2} = 3\).\**3. Compare Limit and Function Value:**\Since \(\lim_{x \to -1} f(x) = 3\) and \(f(-1) = 3\), the function is continuous at \(x = -1\).

Check Continuity at \(x = 10\)

**1. Function Defined at \(x = 10\):**\The piecewise definition applies \(f(x) = \frac{x^2 + 5x + 4}{x^2 + 3x + 2}\) for \(x eq -1\), which includes \(x = 10\).\**2. Calculate the Limit as \(x \to 10\):**\Use the simplified form \(f(x) = \frac{x + 4}{x + 2}\) from Step 2. Substitute directly: \(\lim_{x \to 10} \frac{10 + 4}{10 + 2} = \frac{14}{12} = \frac{7}{6}\).\**3. Compare Limit and Function Value:**\Since \(f(10) = \frac{14}{12} = \frac{7}{6}\) and \(\lim_{x \to 10} f(x) = \frac{7}{6}\), the function is continuous at \(x = 10\).

Key concepts

Limits

Limits are a fundamental concept in calculus that help us understand the behavior of a function as it approaches a certain point. In simple terms, a limit examines what value a function approaches as the input gets closer and closer to a specific number.
Limits are essential for evaluating the continuity of functions, especially at points where direct calculation might be tricky or impossible. They allow us to probe the surroundings of the point and deduce the function's behavior without actually needing the function to be defined at that point itself.
Taking the limit for the function given in the exercise, we consider both the numerator and the denominator separately. By factoring them, we can eliminate common factors and make the limit easier to evaluate. For example, when finding \(\lim_{x \to -1} \frac{x^2 + 5x + 4}{x^2 + 3x + 2} \), factored as \(\frac{(x + 1)(x + 4)}{(x + 1)(x + 2)} \), we cancel out the common term \(x + 1\), simplifying the expression to \(\lim_{x \to -1} \frac{x + 4}{x + 2} \). By substitution of \(x = -1\), the limit computes to 3.
This careful analysis of limits is key to understanding and proving continuity.

Piecewise Functions

Piecewise functions are defined using different expressions based on the input value. Each expression applies to a specific interval or set of intervals in the function's domain. Such functions can often appear disjointed visually, though sometimes they may still be continuous.
In our exercise, the function is given as piecewise:

• For \(x eq -1\), the function follows \(f(x)=\frac{x^{2}+5x+4}{x^{2}+3x+2}\).
• For \(x = -1\), it is defined explicitly as \(f(-1) = 3\).

This method of defining allows for flexibility, especially when certain values might cause complex behavior, such as division by zero. However, piecewise functions require careful examination at the boundaries where the rules switch.
When assessing these functions for continuity, it is vital to verify that: the expressions are defined at critical points, they provide a limit, and the limit matches the function’s actual value if directly called for at that point. Thus, they challenge our analytical skills and prompt us to investigate further into the nature of the function at specific points.

Evaluating Continuity

To evaluate the continuity of functions at particular points, one needs a systematic approach. Continuity at a point \(x = c\) entails three critical checks:

• The function must be defined at \(x = c\).
• The limit as \(x\) approaches \(c\) must exist.
• The limit and the actual function value must be equal: \(\lim_{x \to c} f(x) = f(c)\).

If all these conditions are fulfilled, the function is continuous at that point. For example, in the given exercise at \(x = -1\), \(f(-1)=3\) is defined. The limit \(\lim_{x \to -1} \frac{x+4}{x+2}=3\) is calculated by simplifying the function first to avoid divisions by zero. Lastly, we confirm the function value corresponds to the limit value at this specific point. Therefore, the function is continuous at \(x = -1\).
This evaluation technique provides clarity and ensures that a function behaves predictably as inputs close in on a particular point. It is a crucial skill that underscores much of what we study in calculus.