Sketch the Graph of f(x) = tan 2x

Question:

Sketch the graph of the following function :

\[ f(x)=\tan2x \]

Solution:

We know that

\[ \tan\theta=\frac{\sin\theta}{\cos\theta} \]

Therefore

\[ f(x)=\tan2x \]

The tangent graph increases from \(-\infty\) to \(+\infty\) between consecutive asymptotes.

Whenever

\[ \cos2x=0 \]

the function becomes undefined.

Thus vertical asymptotes occur at

\[ 2x=\frac{\pi}{2}+n\pi \Rightarrow x=\frac{\pi}{4}+\frac{n\pi}{2} \]

Important properties:

  • Period \(=\dfrac{\pi}{2}\)
  • Vertical asymptotes at \(x=\dfrac{\pi}{4}+\dfrac{n\pi}{2}\)
  • The graph increases continuously in each interval

Now calculate some important points:

\[ \begin{aligned} x=-\frac{\pi}{8} &\Rightarrow y=\tan\left(-\frac{\pi}{4}\right)=-1\\[8pt] x=0 &\Rightarrow y=\tan0=0\\[8pt] x=\frac{\pi}{8} &\Rightarrow y=\tan\frac{\pi}{4}=1 \end{aligned} \]

Thus the graph passes through the points

\[ \left(-\frac{\pi}{8},-1\right),\quad (0,0),\quad \left(\frac{\pi}{8},1\right) \]

Plot these points and draw smooth tangent curves approaching the vertical asymptotes.

-π/4 -π/8 0 π/8 π/4 1 -1 x y

Hence, the required graph is shown above.

Graph Features:

  • Period \(=\dfrac{\pi}{2}\)
  • Vertical asymptotes at \(x=\dfrac{\pi}{4}+\dfrac{n\pi}{2}\)
  • The graph increases continuously in each interval
  • Zeros occur at \(x=\dfrac{n\pi}{2}\)

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