Transformation of graphs by modulus function
A function like y=f(x) has different elements. We can apply modulus operator to these elements of the function. There are following different possibilities :
1 :
2 :
3 :
4 :
Plotting concept
The most important point about plotting is to understand that application of modifying operator has different interpretation whether it is applied to independent variable “x” or function definition in x like “f(x)” or it is applied to dependent variable “y” or function definition like “f(y)”. There is a difference in the approach to interpretation.
Clearly, modulus operations have different implications for the graph of f(x). In general, every function can be interpreted to be an operator which operates on its argument, which in itself can be variable like “x”, expression like “
When operator is applied to independent variable or function definition, we evaluate operation of the operator on independent variable or function value. Here, interpretation is based on “evaluation” of the expression (independent variable or function definition) and application of operator thereafter. This applies to the transformations enumerated at (i) and (ii) above. Consider for example,
The function of value at any value x=x is first evaluated. Then, modulus of value is calculated. Finally, it is assigned to y as its value.
This basis of interpretation changes when we apply operator to dependent variable “y” or function definition in “y”. Now the basis of interpretation is that of “assigning” a value to a function and then interpreting the assignment. Such is the case with transformations enumerated at (iii) and (iv) above. Consider for example,
In this case, value of function evaluated at x=x is assigned to modulus function. We interpret equality of the modulus function [y] to a value in accordance with modulus definition. In this case, we know that :
From the point of view of construction of plot, for a single positive value of f(x), say f(x)=4, we have two values of dependent variable i.e. -4 or 4. This needs to be considered while plotting |y|=f(x). In the plot, values of y are plotted against values of x. In this particular instant, there are two points (4,4) and (4,-4) on the graph corresponding to one value of independent variable (4).
Modulus function applied to the independent variable
The form of transformation is depicted as :
It can be seen that modulus operator here modifies independent varaible of the function. In other words, it is like changing input to the function in accordance with nature of modulus function. The input to the function is now either zero or positive number. This has the implication that part of the graph y=f(x) corresponding to negative value of x is not present in the graph of y=f(|x|). Rather, negative value of x is passed as positive value to the function. This means that negative value of independent variable x yields function value which is equal to function value obtained for corresponding positive x whose magnitude is same as that corresponding negative x. It implies that we can obtain function value for negative x by taking image of positive x across y-axis. This is image in y-axis.
From the point of construction of the graph of y=f(|x|), we need to modify the graph of y=f(x) as :
1 : remove left half of the graph
2 : take the mirror image of right half of the graph in y-axis
This completes the construction for y=f(|x|).
Example 1
Problem : Draw graph of
Solution : First we draw graph of sinx. In order to obtain the graph of y=sin|x|, we remove left half of the graph and take the mirror image of right half of the graph of in y-axis.
| Modulus operator applied to sine function |
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Example 2
Problem : Draw graph of
Solution : We first draw graph of
| Modulus operator applied to exponential function |
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In order to obtain the graph of
Example 3
Problem : Draw graph of
Solution : The given expression
In order to draw the graph of
| Modulus operator applied to quadratic function |
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Example 4
Problem : Draw graph of function defined as :
Solution : It is clear that we can obtain given function by applying modulus operator to the independent variable of function given here :
This function, in tern, can be obtained by applying shifting modification to the argument of the function given as :
We, therefore, first draw
| Modulus operator applied to rational function |
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Modulus function applied to the function
The form of transformation is depicted as :
It can be seen that modulus operator here modifies the value of the function itself. In other words, it is like changing output of the function in accordance with nature of modulus function. The output of the function is now either zero or positive number. This has the implication that part of the graph y=f(x) corresponding to negative function values is not present in the graph of y=|f(x)|. Rather, negative function value of f(x) is converted to positive function value. This change in the sign of function takes place without changing magnitude of the value. It implies that we can obtain function values, which correspond to negative function value in y=f(x) by taking image of negative function values across x-axis. This is image in x-axis.
From the point of construction of the graph of y=|f(x)|, we need to modify the graph of y=f(x) as :
(i) take the mirror image of lower half of the graph in x-axis
(ii) remove lower half of the graph
This completes the construction for y=|f(x)|.
Example 5
Problem : Draw graph of
Solution : We first draw the graph of
| Modulus operator applied to cosine function |
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Example 6
Problem : Draw graph of
Solution : We first draw graph
| Modulus operator applied to quadratic function |
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Example 7
Problem : Draw graph of
Solution : We first draw graph
| Modulus operator applied to logarithmic function |
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Modulus function applied to dependent variable
The form of transformation is depicted as :
As discussed in the beginning of module, value of function is first calculated for a given value of x. The value so evaluated is assigned to the modulus function |y|. We interpret assignment to |y| in accordance with the interpretation of equality of the modulus function to a value. In this case, we know that :
Clearly, we need to neglect all negative values of f(x). For every positive value of f(x), there are two values of dependent expressions -f(x) and f(x). It means that we need to take image of upper part of the graph across x-axis. This is image in x-axis.
From the point of construction of the graph of |y|=f(x), we need to modify the graph of y=f(x) as :
1 : remove lower half of the graph
2 : take the mirror image of upper half of the graph in x-axis
This completes the construction for |y|=f(x).
Example 8
Problem : Draw graph of
Solution : We first draw the graph of quadratic function given by
| Modulus operator applied to dependent variable |
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Example 9
Problem : Draw graph of
Solution : We first draw the graph of function given by
| Modulus operator applied to dependent variable |
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Modulus function applied to inverse function
The form of transformation is depicted as :
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)|.
Example 10
Problem : Draw graph of
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
| Modulus operator applied to function in dependent variable |
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Examples
Example 11
Problem : Find domain of the function given by :
Solution : The square root gives the condition :
But denominator can not be zero. Hence,
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 |
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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 :
Example 12
Problem :
Determine graphically the points where graphs of
Solution :
The function
On the other hand,
| Intersection points |
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Clearly, there are three intersection points as shown by solid circles.
Exercises
Exercise 1
Draw the graph of function given by :
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.
Solution 1
| Transformation by modulus operator |
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Exercise 2
2. Draw the graph of function given by :
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.
Solution 2
| Transformation by modulus operator |
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Last edited by Sunil Kumar Singh on Aug 19, 2008 5:20 am GMT-5.