Problem 36
Question
Removable and Nonremovable Discontinuities In Exercises \(35-60,\) find the \(x\) -values (if any) at which \(f\) is not continuous. Which of the discontinuities are removable? \ $$ f(x)=\frac{4}{x-6} $$
Step-by-Step Solution
Verified Answer
The function \( f(x) = \frac{4}{x-6} \) is not continuous at \(x = 6\). The discontinuity at this point is non-removable.
1Step 1: Identify the Denominator
Look at the denominator of the fraction in our function \( f(x) = \frac{4}{x-6} \). The denominator is \(x-6\). Our function is undefined when the denominator is equal to zero.
2Step 2: Solve for x
Determine the x-values that make the denominator zero. For this, we simply set the denominator equal to zero and solve: \(x - 6 = 0\). Adding 6 to both sides gives: \(x = 6\). Therefore, \(x = 6\) makes the denominator zero.
3Step 3: Determine Removability of Discontinuity
As for determining the removability of the discontinuity, we can do so by looking at the limiting value of the function as \(x\) approaches the point of discontinuity from both sides. Since both the limits will approach infinity, this is a non-removable discontinuity. In removable discontinuity, the function approaches a certain value when \(x\) tends to the point of discontinuity.
Key Concepts
Removable DiscontinuityNonremovable DiscontinuityLimits of FunctionsIdentifying Discontinuities
Removable Discontinuity
Understanding a removable discontinuity can feel like joining two puzzle pieces that almost fit together: They're meant to be united but there's a tiny gap. In mathematics, removable discontinuity occurs when a function is not defined at some point, yet the limit exists as it approaches that point.
Imagine you're graphing a function and suddenly you need to lift your pencil off the paper for just one spot, then continue drawing without any jumps or interruptions. That's similar to a removable discontinuity. Often, this can be 'fixed' or 'removed' simply by redefining the function at that single point. A classic example is the hole in the graph of a function, where if that hole was filled, the function would be continuous again.
Imagine you're graphing a function and suddenly you need to lift your pencil off the paper for just one spot, then continue drawing without any jumps or interruptions. That's similar to a removable discontinuity. Often, this can be 'fixed' or 'removed' simply by redefining the function at that single point. A classic example is the hole in the graph of a function, where if that hole was filled, the function would be continuous again.
Nonremovable Discontinuity
On the other hand, a nonremovable discontinuity is like a bridge out on a road: There's no easy way to smoothly continue without a leap. This occurs in functions where the limit does not exist as we approach the point of discontinuity.
A common type of nonremovable discontinuity is a vertical asymptote, which is what the given exercise with the function f(x) = \(\frac{4}{x-6}\) presents. Here, at x = 6, the graph of the function doesn't approach a finite value but instead heads off towards infinity. So there's no 'small fix' to connect the function values, making the discontinuity nonremovable.
A common type of nonremovable discontinuity is a vertical asymptote, which is what the given exercise with the function f(x) = \(\frac{4}{x-6}\) presents. Here, at x = 6, the graph of the function doesn't approach a finite value but instead heads off towards infinity. So there's no 'small fix' to connect the function values, making the discontinuity nonremovable.
Limits of Functions
The concept of limits is pivotal in calculus and involves predicting the values that a function approaches as the input gets closer to a certain point. It's like watching a butterfly flit towards a flower — you're almost certain where it will land, but you can't be entirely sure until it does.
In our exercise context, the limit forms the core tool in determining the type of discontinuity. If the limit of f(x) as x approaches a particular point exists and is finite, we are dealing with a removable discontinuity. If the limit does not exist or is infinite, like the case of f(x) = \(\frac{4}{x-6}\) as x approaches 6, the discontinuity is nonremovable.
In our exercise context, the limit forms the core tool in determining the type of discontinuity. If the limit of f(x) as x approaches a particular point exists and is finite, we are dealing with a removable discontinuity. If the limit does not exist or is infinite, like the case of f(x) = \(\frac{4}{x-6}\) as x approaches 6, the discontinuity is nonremovable.
Identifying Discontinuities
Identifying discontinuities is a critical skill in calculus, akin to a detective finding clues and piecing together what happened at a crime scene. To spot where a function takes an unexpected turn, we search for points where the function behaves oddly — it could jump, halt, or dash off to infinity.
The investigation starts with scrutinizing the function's formula. Points where the function is undefined, like where a denominator is zero, are key places to check. Next, we apply the limits tool to determine the function's behavior approaching these points from both directions. This approach not only reveals where discontinuities lie but also categorizes them as removable or nonremovable.
The investigation starts with scrutinizing the function's formula. Points where the function is undefined, like where a denominator is zero, are key places to check. Next, we apply the limits tool to determine the function's behavior approaching these points from both directions. This approach not only reveals where discontinuities lie but also categorizes them as removable or nonremovable.
Other exercises in this chapter
Problem 35
Finding a \(\delta\) for a Given \(\varepsilon\) In Exercises \(33-36\) , find the limit \(L\) . Then find \(\delta>0\) such that \(|f(x)-L|
View solution Problem 35
Finding a Limit of a Trigonometric Function In Exercises \(27-36,\) find the limit of the trigonometric function. $$ \lim _{x \rightarrow 3} \tan \left(\frac{\p
View solution Problem 36
Finding a One-Sided Limit In Exercises \(33-48,\) find the one-sided limit (if it exists.). $$ \lim _{x \rightarrow 2^{-}} \frac{x^{2}}{x^{2}+4} $$
View solution Problem 36
Finding a \(\delta\) for a Given \(\varepsilon\) In Exercises \(33-36\) , find the limit \(L\) . Then find \(\delta>0\) such that \(|f(x)-L|
View solution