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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|).

Problem : Draw graph of y = sin | x | .

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

Modulus operator applied to the argument of sine function.

Problem : Draw graph of y = e | x + 1 | .

Solution : We first draw graph of y = e x . Then, we shift the graph left by 1 unit to obtain the graph of e x + 1 . At x = 0, y = e 0 + 1 = e . In order to obtain the graph of y = e | x + 1 | , we remove left part of the graph and take the mirror image of right half of the graph of y = e x + 1 in y-axis.

Modulus operator applied to exponential function

Modulus operator applied to the argument of exponential function.

In order to obtain the graph of y = e | x + 1 | , we remove left part of the graph and take the mirror image of right half of the graph of y = e x + 1 in y-axis.

Problem : Draw graph of y = x 2 2 | x | 3

Solution : The given expression f x = x 2 2 | x | 3 is obtained by taking modulus of the independent variable of the corresponding quadratic polynomial in x as given here, f x = x 2 - 2 x - 3 . Hence, we first draw f x = x 2 - 2 x - 3 . The corresponding quadratic equation f x = x 2 - 2 x - 3 = 0 has real roots -1 and 3. The co-efficient of “ x 2 ” is positive. Hence, its plot is a parabola which opens upward and intersects x-axis at x=-1 and x=3.

In order to draw the graph of f x = | x | 2 2 | x | 3 = x 2 2 | x | 3 , we remove left half of the graph and take the mirror image of right half of the core graph of quadratic function in y-axis.

Modulus operator applied to quadratic function

Modulus operator applied to the quadratic function.

Problem : Draw graph of function defined as :

y = 1 | x | + 1

Solution : It is clear that we can obtain given function by applying modulus operator to the independent variable of function given here :

y = 1 x + 1

This function, in tern, can be obtained by applying shifting modification to the argument of the function given as :

y = 1 x

We, therefore, first draw f x = 1 / x . Then we draw g x = f x + 1 = 1 / x + 1 by shifting the graph left by 1 unit. Finally, we draw h x = g | x | = 1 / | x | + 1 by removing left half of the graph and taking mirror image of right half of the graph in y-axis. .

Modulus operator applied to rational function

Modulus operator applied to the argument of rational function.

Modulus function applied to the function

The form of transformation is depicted as :

y = f x y = | f x |

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.

Questions & Answers

how do you translate this in Algebraic Expressions
linda Reply
why surface tension is zero at critical temperature
Shanjida
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what is biological synthesis of nanoparticles
Sanket Reply
what's the easiest and fastest way to the synthesize AgNP?
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Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
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Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
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
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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What is power set
Satyabrata Reply
Period of sin^6 3x+ cos^6 3x
Sneha Reply
Period of sin^6 3x+ cos^6 3x
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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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