Sketch the Graph of f(x) = cot²x
Question:
Sketch the graph of the following function :
\[ f(x)=\cot^2x \]
Solution:
We know that
\[ \cot^2x=(\cot x)^2 \]
Since square of cotangent is always non-negative, the graph always lies above the x-axis.
Whenever
\[ \sin x=0 \]
the cotangent function becomes undefined.
Thus vertical asymptotes occur at
\[ x=n\pi \]
Important properties:
- Period \(=\pi\)
- Range \(y\ge0\)
- Vertical asymptotes at \(x=n\pi\)
Now calculate some important points:
\[ \begin{aligned} x=\frac{\pi}{4} &\Rightarrow y=\cot^2\frac{\pi}{4}=1\\[8pt] x=\frac{\pi}{2} &\Rightarrow y=\cot^2\frac{\pi}{2}=0\\[8pt] x=\frac{3\pi}{4} &\Rightarrow y=\cot^2\frac{3\pi}{4}=1 \end{aligned} \]
Thus the graph passes through the points
\[ \left(\frac{\pi}{4},1\right),\quad \left(\frac{\pi}{2},0\right),\quad \left(\frac{3\pi}{4},1\right) \]
The pattern repeats after every interval
\[ \pi \]
Plot these points and draw smooth curves approaching the vertical asymptotes.
Hence, the required graph is shown above.
Graph Features:
- Period \(=\pi\)
- Range \(y\ge0\)
- Vertical asymptotes at \(x=n\pi\)
- The graph always lies above the x-axis
- The curve approaches infinity near the asymptotes