Problem 30

Question

In Problems , graph each function and, on the basis of the graph, guess where the function is not differentiable. (Assume the largest possible domain.)$$ y=\frac{1}{2+x} $$

Step-by-Step Solution

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Answer

The function is not differentiable at $ x = -2 $.

1Step 1: Understand the Function

We start by identifying the function provided, which is $ y = \frac{1}{2+x} $. This fraction shows that the function is undefined where the denominator equals zero. We need to consider this when determining the domain.

2Step 2: Find Domain

The domain of the function is restricted by the denominator of the fraction. To find the domain, we set the expression in the denominator equal to zero and solve for $ x $: $ 2+x = 0 $. Solving gives $ x = -2 $. Therefore, the domain is all real numbers except $ x = -2 $.

3Step 3: Plot the Graph

With the domain in mind, plot the graph of the function. It will be a hyperbola, with a vertical asymptote at $ x = -2 $ because the function approaches infinity as $ x $ gets closer to $-2$ from either side.

4Step 4: Identify Points of Non-differentiability

Non-differentiability usually occurs at sharp corners, endpoints, or along vertical asymptotes on a graph. Based on our graph, the function does not exist at $ x = -2 $, and thus, it is not differentiable at that point. Hence, the function is not differentiable only at $ x = -2 $.

Key Concepts

Domain of a functionGraphing functionsVertical asymptote

Domain of a function

The domain of a function is a fundamental concept in mathematics. Simply put, it consists of all the possible input values (or 'x' values) for which the function is defined. Take, for example, the function $ y = \frac{1}{2+x} $. Here the denominator, $ 2+x $, plays a critical role in defining the domain. If the denominator is zero, the function becomes undefined.

To find the domain of this function, we solve the equation $ 2+x=0 $

Subtract 2 from both sides to get $ x = -2 $

This solution tells us that the function is undefined when $ x = -2 $. Hence, the domain is all real numbers except $ x = -2 $. Mathematically, we can express the domain as:

$( -\infty, -2) \cup (-2, \infty) $

This means the function can take any real number as input except $-2$. Understanding the domain is crucial, as it guides us on where the function can be plotted on a number line.

Graphing functions

Graphing functions is a very visual way to understand how a function behaves. For the provided function $ y = \frac{1}{2+x} $, graphing helps us visualize its curve and any critical points, like where it suddenly turns or becomes undefined.

When graphing, keep in mind:

The domain: Here, the domain excludes $ x = -2 $.
The asymptotes: The function approaches infinity as $ x $ edges closer to $-2$.

To graph this specific function:

Plot several points by choosing $ x e -2 $.
Observe the curving nature of the hyperbola formed.
Notice the gap and vertical asymptote where $ x = -2 $.

By plotting, it becomes visible how the function stretches towards infinity or negative infinity as it approaches the undefined part of the domain. Graphing serves as an essential tool for recognizing peculiar function behaviors that equations alone may not fully reveal.

Vertical asymptote

A vertical asymptote is a line that a graph of a function approaches but never actually touches or crosses. It's an important concept when dealing with rational functions like $ y = \frac{1}{2+x} $.

For this function, the vertical asymptote appears at the point where the denominator equals zero, which is $ x = -2 $. Here's how it happens:

As $ x $ gets closer to $-2$ from both sides, the function's value heads towards infinity (or negative infinity).
The result is a steep curve that mirrors an invisible boundary at $ x = -2 $.

Vertical asymptotes are essential to recognize on graphs because they denote impenetrable boundaries where the function is undefined. They also indicate points of non-differentiability. This means the function can’t have a tangent line at $ x = -2 $, making it an instant point of disinterest for differentiability discussions. Understanding vertical asymptotes helps in predicting the behavior of graphs at critical points.

Problem 30

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