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y = f x x = | f y |

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 = | cosec y | ; x { - π / 2, π / 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 = | cosec y | .

Modulus operator applied to function in dependent variable

Modulus operator applied to function in dependent variable.

Examples

Problem : Find domain of the function given by :

f x = 1 | sin x | + sin x

Solution : The square root gives the condition :

| sin x | + sin x 0

But denominator can not be zero. Hence,

| sin x | + sin x > 0

| sin x | > 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.

Domain of function

Domain of function is evaluated by comparing transformed graphs.

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 2 n π , 2 n + 1 π

Problem : Determine graphically the points where graphs of | y | = log e | x | and x - 1 2 + y 2 4 = 0 intersect each other.

Solution : The function | y | = log e | x | is obtained by transforming y = log e x . To draw y = 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 | = log e | x | , we transform the graph of y = log e | x | . For this, we remove the lower half and take image of upper half in x-axis.

On the other hand, x - 1 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.

Intersection points

Intersection points are graphically determined.

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

Exercises

Draw the graph of function given by :

f x = 1 [ x ] 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.

Transformation by modulus operator

Transformed graph is shown.

2. Draw the graph of function given by :

f x = | | 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.

Transformation by modulus operator

Transformed graph is shown.

Questions & Answers

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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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Source:  OpenStax, Functions. OpenStax CNX. Sep 23, 2008 Download for free at http://cnx.org/content/col10464/1.64
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