Question

Determine the value of the constant \(a\) for which the function $$f(x)=\left\\{\begin{array}{ll}\frac{x^{2}+3 x+2}{x+1} & \text { if } x \neq-1 \\\a & \text { if } x=-1\end{array}\right.$$ is continuous at -1.

Short answer

Answer: a = 1

Step-by-step solution

Write down the definition of a continuous function

A function is continuous at a point x = c if the following three conditions are met: 1. The function is defined at x = c. 2. The limit of the function as x approaches c exists. 3. The limit of the function as x approaches c is equal to the function's value at x = c, that is, \(\lim_{x\to c} f(x) = f(c)\) In this case, we want the function to be continuous at x = -1.

Find the limit of the function as x approaches -1

We'll find \(\lim_{x\to -1} \frac{x^2 + 3x + 2}{x + 1}\). Notice that the expression can be factored as follows: \(x^2 + 3x + 2 = (x + 1)(x + 2)\) Now, rewrite the expression as $$\lim_{x\to -1} \frac{(x + 1)(x + 2)}{x + 1}$$ As we approach -1, we can cancel out the (x + 1) terms, which will simplify the expression: $$\lim_{x\to -1} (x + 2)$$ Finally, find the limit as x approaches -1: $$\lim_{x\to -1} (x + 2) = (-1) + 2 = 1$$

Set the limit equal to the function's value at x = -1

In order for the function to be continuous at x = -1, we must have \(\lim_{x\to -1} f(x) = f(-1)\). We have already found that the limit is equal to 1. According to the problem, we also know that \(f(-1) = a\). This means that: $$1 = a$$

Conclusion

In order for the function to be continuous at x = -1, the constant a must be equal to 1. Thus, the function becomes: $$f(x)=\left\\{\begin{array}{ll}\frac{x^{2}+3 x+2}{x+1} & \text { if } x \neq-1 \\\1 & \text { if } x=-1\end{array}\right.$$

Key concepts

Limit

When dealing with continuous functions, the concept of a limit is essential to determine if a function behaves predictably around a specific point, even if it isn't defined at that point. For a function to be continuous at a point \(x = c\), the limit as \(x\) approaches \(c\) must exist, and the function must be defined at that point. In our exercise, the function is defined for all points except \(x = -1\), where the piecewise definition provides the value of the function.To find the limit of \(f(x)\) as \(x\) approaches \(-1\), we focused on the expression \(\frac{x^2 + 3x + 2}{x + 1}\). Simplifying the expression, we find that as \(x\) gets close to \(-1\), the terms behave in such a way that their limit can be calculated, even though plugging in \(-1\) would directly make the denominator zero. This key realization allows us to "cancel out" terms and find a limit of 1 at \(x = -1\). This matches the need for \(f(x)\) to be continuous, provided we set the constant value of \(a = 1\) at \(x = -1\).

Piecewise function

A piecewise function is a function that is defined by different expressions, depending on the interval of the input value. These types of functions can be particularly tricky because they may have different rules for continuity at the changeover points, where one piece ends, and another begins.In our function, we have:

• For all \(x eq -1\), \(f(x) = \frac{x^2 + 3x + 2}{x + 1}\)
• At \(x = -1\), \(f(x) = a\)

This illustrates that, while the function is defined by a polynomial expression almost everywhere, the definition changes at \(x = -1\) to adapt to the piecewise rule, ensuring continuity by adjusting the value of \(a\).To ensure continuity, both pieces must "meet" at \(x = -1\), which requires setting \(a\) to the limit of the other part of the function as \(x\) approaches \(-1\). By doing this, we are ensuring the function has no jump, so \(a = 1\) in this context.

Factoring

Factoring is an algebraic process used to simplify expressions or equations, making them easier to evaluate or solve. It involves breaking down complex expressions into products of simpler ones. This is a vital technique when calculating limits, especially when trying to cancel out terms in fractions.In the given exercise, we saw the polynomial numerator \(x^2 + 3x + 2\) that can be factored into \((x + 1)(x + 2)\). This factorization helps us tremendously by enabling us to simplify the expression:\[\frac{x^2 + 3x + 2}{x + 1} = \frac{(x + 1)(x + 2)}{x + 1}\]Upon factoring, the \(x + 1\) terms cancel each other out for \(x eq -1\), simplifying the fraction to \(x + 2\). This simplification is crucial because even though the original fraction is undefined at \(x = -1\) due to division by zero, the simplified expression \(x + 2\) is defined and can thus be evaluated directly, allowing us to find the limit \(\lim_{x \to -1} (x + 2) = 1\). This approach is what allows us to define \(a\) as 1 to ensure continuity at \(x = -1\). Factoring, therefore, not only simplifies expressions but is also fundamental in evaluating limits, as seen in this problem.