Problem 42
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? \ $$ f(x)=\cos \frac{\pi x}{2} $$
Step-by-Step Solution
Verified Answer
The function \(f(x) = \cos(\frac{\pi x}{2})\) is continuous for all x-values and there are no removable or nonremovable discontinuities.
1Step 1: Identifying Continuity of the Function
The function given is \(f(x) = \cos(\frac{\pi x}{2})\), which is the cosine function that has been stretched out by a factor of \( \frac{\pi}{2} \). We know that cosine functions are continuous for all real numbers. Therefore, this function is continuous for all \(x\), because any \(x\) value we substitute into the function will yield a valid real number as output.
2Step 2: Checking for Discontinuities
The function doesn't have any expressions or operations that would result in a undefined output for certain x-values. The function \(f(x)\) is simply a cosine function stretched in the x-direction, and there are no divisions by zero or square roots of negative numbers. Consequently, no \(x\) values would make this function undefined.
3Step 3: Identifying Removable Discontinuities
For a function of real numbers to have removable discontinuities, there must exist certain \(x\)-values that make the function undefined, but the limits from both the left and right at those undefined points would be equal. In the case of \(f(x) = \cos(\frac{\pi x}{2})\), there are no such \(x\)-values that would make the function undefined, let alone satisfy the conditions for removable discontinuities.
Key Concepts
Removable DiscontinuitiesNonremovable DiscontinuitiesTrigonometric FunctionsCosine Function
Removable Discontinuities
A removable discontinuity occurs when there is a hole in the graph of a function. This happens at a certain point if a function is not defined at that point but could be made continuous by appropriately defining or redefining the function at that particular point.
In simpler terms, you can "remove" the discontinuity by filling in a single point on the graph. This is usually caused by a factor that can cancel in the numerator and denominator of a fraction. For instance, something like \( x^2 - a^2 \) in both the numerator and the denominator where \( x = a \) creates this kind of discontinuity.
However, in the function \( f(x) = \cos(\frac{\pi x}{2}) \), there are no fractions or any undefined operations. Hence, there aren't any removable discontinuities. Every point is smoothly connected on the graph without any gaps or holes.
In simpler terms, you can "remove" the discontinuity by filling in a single point on the graph. This is usually caused by a factor that can cancel in the numerator and denominator of a fraction. For instance, something like \( x^2 - a^2 \) in both the numerator and the denominator where \( x = a \) creates this kind of discontinuity.
However, in the function \( f(x) = \cos(\frac{\pi x}{2}) \), there are no fractions or any undefined operations. Hence, there aren't any removable discontinuities. Every point is smoothly connected on the graph without any gaps or holes.
Nonremovable Discontinuities
Nonremovable discontinuities are points on a graph where a function simply cannot be made continuous. These discontinuities can arise from jumps, infinite discontinuities, or oscillations in the graph.
Unlike removable discontinuities, you cannot easily "fix" them without fundamentally changing the function. An example would be where the limit from the left does not equal the limit from the right, or when the limits don't exist at all due to infinite behavior, like in a vertical asymptote.
For \( f(x) = \cos(\frac{\pi x}{2}) \), there are no jumps or infinite behaviors because the cosine function inherently makes use of a continuous wave. Therefore, there are no nonremovable discontinuities present in this function.
Unlike removable discontinuities, you cannot easily "fix" them without fundamentally changing the function. An example would be where the limit from the left does not equal the limit from the right, or when the limits don't exist at all due to infinite behavior, like in a vertical asymptote.
For \( f(x) = \cos(\frac{\pi x}{2}) \), there are no jumps or infinite behaviors because the cosine function inherently makes use of a continuous wave. Therefore, there are no nonremovable discontinuities present in this function.
Trigonometric Functions
Trigonometric functions are mathematical functions that relate the angles of a triangle to the lengths of its sides. The most common trigonometric functions are sine, cosine, and tangent.
These functions are periodic, meaning they repeat their values in regular intervals. This periodic nature makes them continuously oscillating and therefore typically continuous over their defined periods.
In mathematics, trigonometric functions are vital in fields such as physics, engineering, and even in finance, due to their cyclic properties. They model naturally occurring cycles such as waves and rotations.
These functions are periodic, meaning they repeat their values in regular intervals. This periodic nature makes them continuously oscillating and therefore typically continuous over their defined periods.
In mathematics, trigonometric functions are vital in fields such as physics, engineering, and even in finance, due to their cyclic properties. They model naturally occurring cycles such as waves and rotations.
Cosine Function
The cosine function, represented as \( \cos(x) \), is one of the fundamental trigonometric functions. It describes the x-coordinate of a point on the unit circle as a function of the angle \( x \).
One defining feature of the cosine function is that it is an even function, meaning \( \cos(-x) = \cos(x) \). It has a period of \( 2\pi \), after which it repeats its values. This makes cosine inherently continuous across all real numbers.
When you stretch the argument of the cosine function, as in \( \cos(\frac{\pi x}{2}) \), the period changes accordingly, but the continuity principle remains intact. The stretching does not introduce any discontinuities, retaining the same smooth wave-like pattern of the graph.
One defining feature of the cosine function is that it is an even function, meaning \( \cos(-x) = \cos(x) \). It has a period of \( 2\pi \), after which it repeats its values. This makes cosine inherently continuous across all real numbers.
When you stretch the argument of the cosine function, as in \( \cos(\frac{\pi x}{2}) \), the period changes accordingly, but the continuity principle remains intact. The stretching does not introduce any discontinuities, retaining the same smooth wave-like pattern of the graph.
Other exercises in this chapter
Problem 41
Using the \(\varepsilon-\delta\) Definition of Limit In Exercises \(37-48\) , find the limit \(L\) . Then use the \(\varepsilon-\delta\) definition to prove tha
View solution Problem 41
Finding a Limit In Exercises \(41-46,\) write a simpler function that agrees with the given function at all but one point. Then find the limit of the function.
View solution Problem 42
Finding a One-Sided Limit In Exercises \(33-48,\) find the one-sided limit (if it exists.). $$ \lim _{x \rightarrow 3^{+}}\left(\frac{x}{3}+\cot \frac{\pi x}{2}
View solution Problem 42
Using the \(\varepsilon-\delta\) Definition of Limit In Exercises \(37-48\) , find the limit \(L\) . Then use the \(\varepsilon-\delta\) definition to prove tha
View solution