Problem 2 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(1-10,\) find the points of continuity and the points of discontinuity of the function. Identify each type of discontinuity. $$y=\frac{x+1}{x^{2}-4 x+3}$$

Step-by-Step Solution

Answer

The function \(y=\frac{x+1}{x^{2}-4 x+3}\) has discontinuities at x = 1 and x = 3, and these are non-removable or jump discontinuities. The function is continuous for all other real numbers.

Step 1: Identify the Denominator and Set It Equal to Zero

Firstly, identify the denominator in the function \((x^{2}-4 x+3)\), and solve the equation by setting the denominator equal to zero to find the values of x where the function does not exist.

Step 2: Solve the Equation for x

Next, solve the equation \(x^{2}-4 x+3 = 0\) for x, by either factoring, completing the square or using the quadratic formula. After solving the equation, we get x = 1 and x = 3 as the solutions.

Step 3: Identify the Type of Discontinuity

We can determine the type of discontinuity by finding the limit of the function as x approaches the solutions from the left and right side. Also by substituting these values in the function, it's clear neither 1 nor 3 is a removable discontinuity (also called a hole). Since the function is undefined at x = 1 and x = 3, these are points of discontinuity and are called 'non removable' or 'jump' discontinuities. For all other real numbers, the function is continuous.

Key Concepts

Types of Discontinuities

When we talk about functions, it's essential to understand where they behave nicely and where they might act up. Discontinuities are the 'act up' parts, where a function doesn't follow a smooth path. There are several types of discontinuities you should be aware of:

• Removable Discontinuities: These occur at a certain point if the function approaches a particular value, but there's a gap at that point. We can 'remove' this discontinuity by redefining the function at that point.


• Jump Discontinuities: Occur when the left-hand limit and the right-hand limit of the function exist at a certain point but are not equal to each other. The function 'jumps' from one value to another.


• Infinite Discontinuities: These happen when the function heads off to infinity at a certain point, often seen in functions with vertical asymptotes.


• Oscillating Discontinuities: When the function swings back and forth indefinitely as it approaches a certain point, it is called an oscillating discontinuity. It's like not being able to make up its mind!

Once the type of discontinuity is identified, it's much easier to understand how a function behaves and how to tackle it. In the given exercise, we found that the function has 'jump' discontinuities at x = 1 and x = 3, where the function is not defined.

Limit of a Function

The limit of a function is like a destination towards which the function is headed, even if it never actually gets there. It tells us the value the function approaches as the input (or 'x') gets closer to a certain number. Let’s break it down:

Let's say we have a function f(x) and we want to know what happens as x approaches a certain value 'c'. The limit of f(x) as x approaches c is denoted as \(\lim_{x\to c} f(x)\). It's like saying, 'Hey, as we get closer and closer to 'c', what's f(x) getting closer to?'

Calculating limits can involve some detective work:

• Direct substitution: If plugging 'c' into the function doesn’t cause any issues (like dividing by zero), then the limit is just f(c).


• Factoring: Sometimes, you’ve got to factor a polynomial to simplify and find the limit by direct substitution.


• Special techniques: When you have more complex cases, like '0/0', you would use L'Hôpital's Rule or conjugate methods.

In the exercise, to determine the type of discontinuity at x = 1 and x = 3, calculating the limit is key. It showed that the function did not gradually approach a finite value, leading to the identification of 'jump' discontinuities at these points.

Solving Quadratic Equations

Quadratic equations are like puzzles that ask: what x-values make this parabola-shaped curve cross the x-axis? The standard form of a quadratic equation is \(ax^2 + bx + c = 0\), and there are several methods to solve them:

• Factoring: If the equation can be factored, this is often the quickest way to find the roots.


• Completing the Square: This technique involves creating a perfect square trinomial from the quadratic equation, then taking the square root of both sides to solve for x.


• Quadratic Formula: The trusty quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), works every time for any quadratic equation, as long as you're comfortable with the arithmetic involved.

Each method has its own place and usefulness, depending on the nature of the quadratic equation you're dealing with. In the original exercise, solving the equation \(x^2 - 4x + 3 = 0\) was necessary to find the points of discontinuity. Factoring the quadratic polynomial was the method used, which yielded the roots x = 1 and x = 3. These solutions play a crucial role in understanding where the function is not defined and thus discontinuous.