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

Question: Determine the value of constant \(a\) such that the function \(f(x)\) given below is continuous at \(x=-1\): $$ f(x)=\left\\{\begin{array}{ll} \frac{x^{2}+3 x+2}{x+1} & \text { if } x \neq-1 \\\a &\text { if } x=-1\end{array}\right. $$ Answer: The function is continuous at \(x=-1\) when \(a = 1\).

Step-by-step solution

Rewrite the function in a simpler form for \(x \neq -1\).

Since we only need to deal with the function when \(x \neq -1\), let's rewrite the function for this condition by simplifying the expression. For \(x \neq -1\), we have: $$ f(x)=\frac{x^{2}+3 x+2}{x+1} = \frac{(x+1)(x+2)}{x+1} $$ The factor \((x+1)\) on the numerator and denominator cancel out, leaving us with: $$ f(x) = x + 2 $$

Find the limit of the function as \(x\) approaches -1.

Now that we have a simpler expression for \(f(x)\), let's find the limit as \(x\) approaches -1: $$ \lim_{x \to -1} f(x) = \lim_{x \to -1} (x + 2) $$ As this is a linear function, it is continuous everywhere, so the limit is simply the value of the function at \(x = -1\): $$ \lim_{x \to -1} f(x) = -1 + 2 = 1 $$

Set the limit equal to \(a\).

Since we want \(f(x)\) to be continuous at \(x = -1\), the limit from the previous step should be equal to the function value at this point. In other words, we want \(f(-1) = a\): $$ a = \lim_{x \to -1} f(x) = 1 $$

Write the final continuous function.

Now that we have determined the value of \(a\) for which the function is continuous, we can write the continuous function as: $$ f(x)=\left\\{\begin{array}{ll} x + 2 & \text { if } x \neq-1 \\\1 & \text { if } x=-1\end{array}\right. $$ So, the function is continuous at \(x=-1\) when \(a = 1\).

Key concepts

Limits

Understanding the concept of limits is fundamental in calculus and pertains to finding the value that a function approaches as the input approaches a certain point. Limits help in understanding the behavior of functions as they get close to a specific value, without actually reaching that point.

For instance, in the solved exercise, we were looking for the limit of the function as \(x\) approaches -1. It's important to remember that limits can be direct, one-sided (from the left or the right), or they may not exist at all, especially if the function behaves erratically near the point in question.

In simpler cases, like linear functions, finding the limit is straightforward, as these functions are continuous everywhere except where they are undefined. Thus, the limit as \(x\) approaches any value is simply the function's value at that point. The exercise highlighted the coherence of this principle by showing that the limit of \(f(x) = x + 2\) as \(x\) approaches -1 is indeed \(-1 + 2 = 1\).

Piecewise Functions

Piecewise functions are functions defined by multiple sub-functions, each of which applies to a certain interval in the domain. These functions are powerful tools for modeling complex, real-world situations where different conditions call for different responses.

In our exercise, the function \(f(x)\) was defined differently for \(x\) at -1 and at other values. This is a classic example of a piecewise function. To ensure the function is continuous, especially at points where its definition switches between sub-functions, the function's values must align at that point.

Navigating through piecewise functions

When dealing with piecewise functions, it's important to identify the intervals and the corresponding function rules. If these functions have common points, they should provide the same output to maintain continuity. The solution process showed this by setting the limit equal to the value of \(a\) at \(x=-1\), ensuring a seamless transition between the function's different expressions.

Simplifying Expressions

Simplifying expressions in algebra involves reducing them to the simplest form possible, often making it easier to understand and work with the function. In the context of our exercise, the process of simplifying involved factoring and canceling out identical factors on the numerator and denominator of the rational function.

This process not only clarifies the function's form but also is crucial in determining limits and ensuring continuity. By simplifying \(\frac{x^{2}+3x+2}{x+1}\) to \(x+2\), we avoided potential complications at the point \(x=-1\), where the function was initially undefined.

The beauty of simplification

While simplifying, it's vital to look out for common factors and apply basic algebraic operations. This steps reveal the true nature of the function and often, as in the exercise, lead to a more elementary expression, facilitating the understanding and the computation of limits where the original function may be too complex to evaluate directly.