Finding the Continuous Intervals for a function

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SUMMARY

The function $f(x)=tan(2x)$ is continuous everywhere except at its vertical asymptotes, which occur at $x=\frac{n \pi}{4}$ for odd integers n. The vertical asymptotes arise when the cosine function in the denominator equals zero, specifically at points $x=\pm \frac{\pi}{4}, \pm \frac{3\pi}{4}, \ldots$. Thus, the continuous intervals for $f(x)$ can be defined as the ranges between these asymptotes.

PREREQUISITES
  • Understanding of trigonometric functions, specifically tangent and cosine.
  • Knowledge of vertical asymptotes and their implications on function continuity.
  • Familiarity with the unit circle and periodicity of trigonometric functions.
  • Basic algebra skills for manipulating trigonometric equations.
NEXT STEPS
  • Study the properties of the tangent function and its periodicity.
  • Learn how to identify and graph vertical asymptotes in trigonometric functions.
  • Explore the concept of continuity and discontinuity in calculus.
  • Investigate the behavior of $f(x)=tan(kx)$ for different values of k.
USEFUL FOR

Students and educators in mathematics, particularly those focusing on calculus and trigonometry, as well as anyone studying the properties of periodic functions and their graphs.

cbarker1
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I need some help find some continuous intervals for $f(x)=tan(2x)$. I know there are vertical asympotes when x=pi/4+2*pi*n for positive integers. Thank you for your help.

CBarker1
 
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Cbarker1 said:
I need some help find some continuous intervals for $f(x)=tan(2x)$. I know there are vertical asympotes when x=pi/4+2*pi*n for positive integers. Thank you for your help.

CBarker1

$f(x)=tan(2x)$ is continuous everywhere except the vertical asymptotes. To find the vertical asymptotes,we set the denominator $0$.So,as $tan(2x)=\frac{sin(2x)}{cos(2x)}$,we set $cos(2x)=0 \Rightarrow x=\pm \frac{\pi}{4},\pm \frac{3\pi}{4},...$
Therefore,$tan(2x)$ is continuous everywhere except at $x$,where $x=\frac{n \pi}{4}$,where $n$ odd numbers.
 
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