S8.02.4.58 type of discontinuity tikx with 1/sqrt(1-cos(2x))

In summary, the conversation discusses the function $f(x)=\dfrac{1}{\sqrt{1-\cos{2x}}}$ and its graph on desmos. The question is asked about the type of discontinuity at $x=0$ and the left and right limits at that point. The answer is not fully provided but it is mentioned that the limit goes to infinity from both positive and negative directions. There is also mention of using TikX plot code to graph the function on Overleaf. Clarification is requested on the types of discontinuity and the definitions of left and right limits.
  • #1
karush
Gold Member
MHB
3,269
5
$\tiny{s8.2.4 58}$

desmos
02_2_4_58.png.png

Let
$f(x)=\dfrac{1}{\sqrt{1-\cos{2x}}}$
(a) Graph $f$ What type of discontinuity does it appear to have at 0?\\
(b) Calculate the left and right limits of $f$ at 0. \\
Do these valuesn' confirm your answer to part (a)?

doesn't the limit going to 0 infinity both + and -

what is the tikx plot code for
$\dfrac{1}{\sqrt{1-\cos{2x}}}$ {1 / sqrt(x)} is as far as I gotalso what plotted in Overleaf didn't preview here
 
Physics news on Phys.org
  • #2
It's not clear to me what you are asking. What "types" of discontinuity are there? How is each defined?

Do you know what "left" and "right" limits are? (Since they ask about "left" and "right" of x= 0 you might want your graph to include some negative values of x.
 

1. What is the definition of "S8.02.4.58 type of discontinuity tikx with 1/sqrt(1-cos(2x))"?

The "S8.02.4.58 type of discontinuity tikx with 1/sqrt(1-cos(2x))" refers to a specific type of discontinuity in a mathematical function, where the function is undefined at a certain point due to the presence of a square root of the difference between 1 and the cosine of twice the input value.

2. What causes this type of discontinuity to occur?

This type of discontinuity occurs when the input value of the function results in a division by zero, which is undefined. In this case, the division by zero is caused by the square root of the difference between 1 and the cosine of twice the input value.

3. How can this type of discontinuity be identified in a function?

This type of discontinuity can be identified by analyzing the function and looking for any values of the input that would result in a division by zero. In this case, the presence of the square root of the difference between 1 and the cosine of twice the input value is a clear indicator of this type of discontinuity.

4. Can this type of discontinuity be removed or avoided?

In most cases, this type of discontinuity cannot be removed or avoided, as it is a fundamental characteristic of the function. However, in some cases, the function can be simplified or rewritten in a way that avoids this type of discontinuity.

5. What implications does this type of discontinuity have for the function?

This type of discontinuity can have significant implications for the function, as it means that the function is undefined at certain points. This can affect the overall behavior and accuracy of the function, and may require special considerations when working with it.

Similar threads

MHB Where are the one-sided derivatives of f(x) equal?
    • Calculus
Replies
2
Views
713
MHB 5.5.2 average value of sqrt{x} [0,4]
    • Calculus
Replies
1
Views
1K
MHB Finding the Derivative of the Inverse Function of a Cubic Polynomial
    • Calculus
Replies
5
Views
2K
MHB 219 GRE what is the value of g'(1)
    • Calculus
Replies
3
Views
1K
MHB 219 AP Calculus Exam Inverse function
    • Calculus
Replies
1
Views
1K
I Find the directional derivative of ##f## at the given point
    • Differential Equations
Replies
3
Views
677
MHB 3.2.15 mvt - Mean value theorem: graphing the secant and tangent lines
    • Calculus
Replies
3
Views
1K
MHB -7.8.98 amplitude period PS VS graph. of cos eq
    • General Math
Replies
5
Views
950
MHB Q1 Can you pass this 3 question AP calculus Quiz in 10 minutes
    • Calculus
Replies
3
Views
726
MHB Summing Series Limit as N→∞
    • Calculus
Replies
3
Views
942

Hot Threads

Recent Insights

Back
Top