4.2 Transformation of graphs by modulus function  (Page 4/4)

Problem : Draw graph of $x=\left|\mathrm{cosec}y\right|;x\in \{-\pi /\mathrm{2,}\pi /2\}$ .

Solution : The inverse of base function is cosec⁻¹x. We first draw the graph of inverse function. Then, we take mirror image of left half of the graph in y-axis and remove left half of the graph to complete the construction of graph of $x=\left|\mathrm{cosec}y\right|$ .

Problem : Find domain of the function given by :

$$f\left(x\right)=\frac{1}{\sqrt{\left|\sin x\right|+\sin x}}$$

Solution : The square root gives the condition :

$$\Rightarrow \left|\sin x\right|+\sin x\ge 0$$

But denominator can not be zero. Hence,

$$\Rightarrow \left|\sin x\right|+\sin x>0$$

$$\Rightarrow \left|\sin x\right|>-\sin x$$

We shall make use of graphing technique to evaluate the interval of x. Since both functions are periodic. It would be indicative of the domain if we confine our consideration to 1 period of sine function (0, 2π) and then extend the result subsequently to other periodic intervals.

We first draw sine function. To draw |sinx|, we take image of lower half in x-axis and remove the lower half. To draw “–sinx”, we take image of y=sinx in x-axis.

From the graph, we see that |sinx| is greater than “-sinx” in (0,π). Note that end points are not included. The domain is written with general notation as :

$$x\in \left(2n\pi ,\left(2n+1\right)\pi \right)$$

Problem : Determine graphically the points where graphs of $\left|y\right|={\log }_e\left|x\right|$ and ${\left(x-1\right)}^2+y^2-4=0$ intersect each other.

Solution : The function $\left|y\right|={\log }_e\left|x\right|$ is obtained by transforming $y=\log {}_{e}x$ . To draw $y={\log }_e\left|x\right|$ , we need to remove left half (but here there is no left half) and take image of right half in y-axis. To draw $\left|y\right|=\log {}_e\left|x\right|$ , we transform the graph of $y=\log {}_e\left|x\right|$ . For this, we remove the lower half and take image of upper half in x-axis.

On the other hand, ${\left(x-1\right)}^2+y^2-4=0$ is a circle with center at 1,0 having radius of 2 units. Finally, superposing two graphs, we determine the intersection points.

Clearly, there are three intersection points as shown by solid circles.