Sketch the Graph of f(x) = cosec²x
Question:
Sketch the graph of the following function :
\[ f(x)=\cosec^2x \]
Solution:
We know that
\[ \cosec^2x=\frac{1}{\sin^2x} \]
Since square of cosecant is always positive, the graph always lies above the x-axis.
Whenever
\[ \sin x=0 \]
the function becomes undefined.
Thus vertical asymptotes occur at
\[ x=n\pi \]
Important properties:
- Period \(=\pi\)
- Range \(y\ge1\)
- Vertical asymptotes at \(x=n\pi\)
- Minimum value \(=1\)
Now calculate some important points:
\[ \begin{aligned} x=\frac{\pi}{2} &\Rightarrow y=\cosec^2\frac{\pi}{2}=1\\[8pt] x=\frac{\pi}{4} &\Rightarrow y=\cosec^2\frac{\pi}{4}=2\\[8pt] x=\frac{3\pi}{4} &\Rightarrow y=\cosec^2\frac{3\pi}{4}=2 \end{aligned} \]
Thus the graph passes through the points
\[ \left(\frac{\pi}{2},1\right),\quad \left(\frac{\pi}{4},2\right),\quad \left(\frac{3\pi}{4},2\right) \]
Plot these points and draw smooth curves approaching the vertical asymptotes.
Hence, the required graph is shown above.
Graph Features:
- Period \(=\pi\)
- Range \(y\ge1\)
- Vertical asymptotes at \(x=n\pi\)
- The graph always lies above the x-axis
- The curve approaches infinity near the asymptotes