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

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$y=f\left(x\right)\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}x=|f\left(y\right)|$

The invertible function x= f(y) has its inverse function given by y=f⁻¹(x). Alternatively, if a function is defined as y=f⁻¹(x), then variables x and y are related to each other such that x=f(y). We conclude that graph of y=f⁻¹(x) is same as graph of x=f(y) with the same orientation of x and y axes. It is important to underline here that we transform (change) graph of inverse of given function i.e. y=f⁻¹(x) to get the transformation of graph of x=f(y). Further x and y coordinates on the graph correspond to x and y values.

We interpret assignment of |f(y)| to x in the given graph in accordance with the definition of modulus function. Consider x=|f(y)|. But, modulus can not be equated to negative value. Hence, x can not be negative. It means we need to discard left half of the graph of inverse function y=f⁻¹(x). On the other hand, modulus of negative or positive value is always positive. Hence, positive value of x=a correspond to two values of function in dependent variable, a=±f(y). Corresponding to these two function values in y, we have two values of y i.e. f⁻¹(a) and f⁻¹(-a). In order to plot two values, we need to take mirror image of the left half of the graph of y=f⁻¹(x) across y-axis. This is image in y-axis.

From the point of construction of the graph of x=|f(y)|, we need to modify the graph of y=f⁻¹(x) i.e. x=f(y) as :

1 : take mirror image of left half of the graph in y-axis

2 : remove left half of the graph

This completes the construction for x=|f(y)|.

Problem : Draw graph of $x=|\mathrm{cosec}y|;\phantom{\rule{1em}{0ex}}x\in \left\{-\pi /2,\pi /2\right\}$ .

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 $x=|\mathrm{cosec}y|$ .

## Examples

Problem : Find domain of the function given by :

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

Solution : The square root gives the condition :

$⇒|\mathrm{sin}x|+\mathrm{sin}x\ge 0$

But denominator can not be zero. Hence,

$⇒|\mathrm{sin}x|+\mathrm{sin}x>0$

$⇒|\mathrm{sin}x|>-\mathrm{sin}x$

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 $|y|={\mathrm{log}}_{e}|x|$ and ${\left(x-1\right)}^{2}+{y}^{2}-4=0$ intersect each other.

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

## Exercises

Draw the graph of function given by :

$f\left(x\right)=\frac{1}{\left[x\right]-1}$

Hints : Draw 1/x, which is a hyperbola with center at (0,0). Then draw 1/x-1. It is a hyperbola shifted right by 1 unit. Its center is (1,0). Remove left half and take the image of right half in y-axis.

2. Draw the graph of function given by :

$f\left(x\right)=||\frac{1}{x}|-1|$

Hints : Draw 1/x, which is a hyperbola with center at (0,0). Then draw |1/x|. Take image of lower half in x-axis. Remove lower half. To draw |1/x|-1, shift down the graph of |1/x| by 1 unit. To draw ||1/x|-1|, Take image of lower half of the graph of |1/x|-1 in x-axis. Remove lower half.

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in general
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Period of sin^6 3x+ cos^6 3x