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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 :

(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)|.

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

Solution : We first draw the graph of y = cos x . Then, we take the mirror image of lower half of the graph in x-axis and remove lower half of the graph to complete the construction of graph of y = | cos x |

Modulus operator applied to cosine function

Modulus operator applied to cosine function.

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

Solution : We first draw graph y = x 2 2 x 3 . The roots of corresponding quadratic equation are -1 and 3. After plotting graph of quadratic function, we take the mirror image of lower half of the graph in x-axis and remove lower half of the graph to complete the construction of graph of y = | x 2 2 x 3 |

Modulus operator applied to quadratic function

Modulus operator applied to quadratic function.

Problem : Draw graph of y = | log 10 x | .

Solution : We first draw graph y = log 10 x . Then, we take the mirror image of lower half of the graph in x-axis and remove lower half of the graph to complete the construction of graph of y = | log 10 x | .

Modulus operator applied to logarithmic function

Modulus operator applied to logarithmic function.

Modulus function applied to dependent variable

The form of transformation is depicted as :

y = f x | y | = f x

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 :

| y | = f x ; f x > 0 y = ± f x

| y | = f x ; f x = 0 y = 0

| y | = f x ; f x < 0 Modulus can not be equated to negative value. No solution

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).

Problem : Draw graph of | y | = x 1 x 3 .

Solution : We first draw the graph of quadratic function given by y = x 1 x 3 . Then, we remove lower half of the graph and take mirror image of upper half of the graph in x-axis to complete the construction of graph of | y | = x 1 x 3 .

Modulus operator applied to dependent variable

Modulus operator applied to dependent variable.

Problem : Draw graph of | y | = tan - 1 x .

Solution : We first draw the graph of function given by y = tan - 1 x . Then, we remove lower half of the graph and take mirror image of upper half of the graph in x-axis to complete the construction of graph of y = tan - 1 x .

Modulus operator applied to dependent variable

Modulus operator applied to dependent variable.

Modulus function applied to inverse function

The form of transformation is depicted as :

Questions & Answers

a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
how did I we'll learn this
Noor Reply
f(x)= 2|x+5| find f(-6)
Prince Reply
f(n)= 2n + 1
Samantha Reply
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
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
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
Period of sin^6 3x+ cos^6 3x
Sneha Reply
Period of sin^6 3x+ cos^6 3x
Sneha Reply

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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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