Problem 28 - Single Variable Calculus

Question

Let $$ f(x)=\left\\{\begin{array}{ll} \frac{x^{2}-4}{x+2} & \text { if } x \neq-2 \\ k & \text { if } x=-2 \end{array}\right. $$ Find the value of \(k\) that will make \(f\) continuous on \((-\infty, \infty)\).

Step-by-Step Solution

Answer

For the function $f(x)$ to be continuous on $(-\infty, \infty)$, the value of \(k\) should be \(-4\).

Step 1: Simplify the expression

We start by factoring the polynomial in the numerator: \((x^2 - 4) = (x - 2)(x + 2)\). Therefore, the expression can be written as: \[ \frac{x^2 - 4}{x+2} = \frac{(x - 2)(x + 2)}{x + 2} \]

Step 2: Cancel common factors

We now cancel the common factors of the numerator and denominator, which is (x + 2). The expression can then be simplified to: \[ \frac{(x - 2)(x + 2)}{x + 2} = x - 2 \]

Step 3: Find the limit as x approaches -2

Now, we find the limit of the expression \(x - 2\) as x approaches -2: \[ \lim_{x \to -2}(x - 2) = -2 - 2 = -4 \]

Step 4: Set the limit equal to k

Since we want \(f(x)\) to be continuous at \(x = -2\), we need to set the limit equal to the value of k: \[ k = -4 \] Therefore, for the function f to be continuous on \((-\infty, \infty)\), the value of k should be -4.

Key Concepts

Limits

Limits help us understand the behavior of functions as their inputs approach a certain value. Consider a road that appears to disappear over the horizon; you can't see what's beyond, but you can predict where it leads based on the visible stretch. Similarly, evaluating a limit involves predicting the value a function approaches as the input nears a point.

• This is especially useful when a function isn't defined at a point. For example, in our exercise, as we approach x = -2, we need to evaluate the function's behavior because it's not directly defined there.
• The formal process involves finding \( \lim_{x \to a} f(x) \), to predict the output value as x approaches a specific number 'a'.
• When a limit exists and equals the function's value at that point, it helps in establishing continuity.

In practice, calculating limits often involves simplification techniques like factoring to remove any indeterminate forms, enabling smooth prediction of the function's behavior.

Factoring Polynomials

Factoring polynomials means breaking down an expression into simpler components, or factors, that when multiplied give the original expression. Consider it akin to dividing a complex puzzle into smaller, more manageable pieces that fit together perfectly.

• For quadratic polynomials like \( x^2 - 4 \), factoring is often straightforward. We use identities such as \( a^2 - b^2 = (a - b)(a + b) \) to factor it into \( (x - 2)(x + 2) \).
• This skill is essential in calculus, enabling the simplification of expressions, which is crucial in solving limits and identifying function behavior around critical points.
• Once polynomials are factored, you can proceed to simplify expressions by canceling common factors, easing the calculation of limits.

Factoring not only simplifies expressions but also uncovers hidden zeros of polynomials, which are the key to understanding their properties and behaviors.

Canceling Common Factors

Canceling common factors is the act of simplifying fractions by removing identical terms in the numerator and denominator. This is an essential strategy when simplifying expressions, especially when dealing with limits.

• In our exercise, after factoring the numerator of \( \frac{x^2 - 4}{x + 2} \) to \( \frac{(x - 2)(x + 2)}{x + 2} \), the \( (x + 2) \) term appears in both the numerator and denominator.
• By canceling the common factor \((x + 2) \), the expression simplifies to \( x - 2 \), removing potential indeterminate forms and making it easier to evaluate at points like \( x = -2 \).
• This step is crucial for finding the true value of the limit, allowing us to bypass undefined expressions and focus on the function's behavior around specific points.

Understanding and applying common factor cancellation leads to a more streamlined approach in calculus, paving the way for clearer and more accurate limit evaluation.

Continuous Functions

A function is continuous when there is no interruption in its graph; think of drawing the graph without lifting your pencil. Mathematically, a function \( f(x) \) is continuous at a point \( x = a \) if:

• \( f(a) \) is defined, meaning there is an actual value at that point.
• The limit of \( f(x) \) as it approaches \( a \) from both sides exists.
• The limit is equal to \( f(a) \).

In our exercise, to make the function \( f(x) \) continuous on \(( -\infty, \infty )\), it must be continuous at \( x = -2 \). The strategy involves calculating the limit as \( x \) approaches \(-2\), which was \(-4\), and setting \( f(-2) = k \) to this limit value. Thus, \( k = -4 \) ensures no break in continuity.Understanding continuity provides insight into the behavior and integrity of functions, making it a key concept in calculus and mathematical analysis.