Problem 27 - Calculus of a Single Variable

Question

Continuity of a Function In Exercises \(27-30,\) discuss the continuity of each function. $$ f(x)=\frac{1}{x^{2}-4} $$

Step-by-Step Solution

Answer

The function \(f(x) = \frac{1}{x^{2}-4}\) is continuous everywhere except at \(x = 2\) and \(x = -2\).

Step 1: Identify the Denominator

The denominator of the function is \(x^{2}-4\). This is the part of the function where if it equals to zero, the function will be undefined.

Step 2: Solve the Denominator Equal to Zero

To find out the points where the function might not be continuous, set the denominator equal to zero and solve for \(x\):\n\[x^{2}-4=0\]\nThis is a simple quadratic equation and can be solved easily by factoring or using the quadratic formula. Factoring will be the easiest in this case:

Step 3: Factoring the Equation

Factoring \(x^{2} - 4 = 0\) gives us \((x - 2)(x + 2) = 0\).

Step 4: Solve for x

Setting both factors equal to zero gives us the solutions \(x = 2\) and \(x = -2\). This means that the denominator of the function equals zero at \(x = 2\) and \(x = -2\), making the function undefined at these points.

Step 5: Conclusion

Therefore, the function \(f(x) = \frac{1}{x^{2}-4}\) is continuous except at \(x = 2\) and \(x = -2\). At these points, the function is not defined, hence not continuous.

Key Concepts

Undefined Function Points

When we are discussing the continuity of functions, especially rational functions like the example provided, identifying undefined function points is a crucial first step. These are the points where a function does not have a value. In the context of a rational function, this occurs where the denominator is zero because division by zero is undefined in mathematics.
In the given function, to determine when it will be undefined, we need to find the values of the variable, in this case, 'x', that will make the denominator, which is a quadratic expression, equal to zero. The given function, f(x) = \(\frac{1}{x^{2}-4}\), will therefore not be defined when its denominator, \(x^{2}-4\), is zero.
By solving the equation \(x^2 - 4 = 0\), as shown in the step by step solution, the points where the function f(x) is undefined are found to be \(x = 2\) and \(x = -2\). With these findings, we understand that for all other values of 'x', the function is defined, but at \(x = 2\) and \(x = -2\), the function f(x) will not exist, hence introducing breaks in its continuity.

Factoring Quadratic Equations

The process of factoring quadratic equations is a fundamental skill for solving these types of equations. A quadratic equation is an equation of the form \(ax^2 + bx + c = 0\), where 'a', 'b', and 'c' are constants. Factoring involves rewriting the quadratic expression as a product of two binomials. When a quadratic expression is set equal to zero, factoring is a method used to solve for the variable.

Why Factor Quadratics?

• Factoring simplifies the equation, making it easier to find the roots or solutions.
• Provides a clear understanding of the behavior of a quadratic function, such as its zeros or x-intercepts, where the graph of the function touches or crosses the x-axis.

The step by step solution illustrated factoring the denominator \(x^2 - 4\) by recognizing it as a difference of squares, a special factoring case where \(a^2 - b^2 = (a - b)(a + b)\). This simplification is crucial because it allows us to break down the quadratic equation into simpler linear factors, helping to find the values of 'x' that make the original function undefined.

Solving Quadratic Equations

Solving quadratic equations is an essential competence in algebra. These equations can usually be solved either by factoring, completing the square, or using the quadratic formula. In the scenario at hand, the quadratic equation \(x^2 - 4 = 0\) was solved by factoring.
When factoring a quadratic equation, once broken down into \((x - a)(x - b) = 0\), we apply the Zero Product Property, which states if the product of two factors is zero, then at least one of the factors must be zero. Here, the factors \((x - 2)\) and \((x + 2)\) lead us to the solutions \(x = 2\) and \(x = -2\).

Alternate Methods

• Completing the square: This involves rearranging the equation to form a perfect square trinomial.
• Quadratic formula: This is the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), used when factoring is not feasible or more complicated.

Each method has its advantages, depending on the complexity and form of the quadratic equation in question. In classroom settings and homework problems like the one we're examining, recognizing the best approach is a key analytical skill. This enhances not just problem-solving efficiency but also a deeper understanding of algebraic structures and their properties.