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Working rules

A. Writing composition :

  • Write fog(x) = f(y), where y = g(x)

B. Finding domain of the composition :

  • Write down the domain interval of argument function “g(x)” of the composition.
  • Write the domain interval of the main function “f(x)” by substituting independent variable by the argument function “g(x)” itself.
  • Interpret the interval with argument function, “g(x)”.
  • Intersection of two intervals is the valid interval of composition.

This exercise module did not follow immediately after the module on composition of functions. We needed to know different function types first to apply the concept with them.

Problem 1: A function f(x) is given as :

f x = { a x n } 1 n

where a>0, x>0 and "n" is a positive integer. Find f{f(x)}

Solution :

Statement of the problem : The domain of the given function is positive number as x>0. In order to find, the composition, we evaluate f(y), where y = f(x).

f { f x } = f y = a y n 1 n = [ a { a x n 1 / n } n ] 1 / n

f { f x } = a a x n 1 n = x n 1 n = x ; x > 0

Here, composition is that of function with itself. As such, domain of composition is equal to intersection of domain of the given function with itself. But, the intersection of an interval with itself is same interval. Hence, we have retained the domain interval of the composition same as that of given function.

Problem 2: A function f(x) is given as :

f x = { 2 x n } 1 n

where x>0 and "n" is a positive integer. Prove that :

f { f x } + f { f 1 x } 2

Solution :

Statement of the problem : The domain of the given function is positive number as x>0. In order to prove the inequality, we need to determine each composition on the left hand side of the given inequality.

We have seen in earlier example that if f x = { a x n } 1 n , then f{f(x) = x. Hence, if f x = { 2 x n } 1 n , then f{f(x) = x. Similarly, we determine f{f(1/x)}. Here,

f { f 1 x } = f y = [ a { a 1 x n 1 / n } n ] 1 / n = a a 1 x n 1 / n

f { f 1 x } = 1 x n 1 n = 1 x

Substituting these values in the LHS of the inequality, we have :

LHS = f { f x } + f { f 1 x } = x + 1 x

Using algebraic identity a 2 + b 2 = a b 2 + 2 a b , we have :

LHS = x 1 x 2 + 2

But, the square term is a non-negative number. Hence,

LHS 2

f { f x } + f { f 1 / x } 2

Problem 3: A function is defined as :

| -1 ; -2≤x≤0 f(x) = || x-1 ; 0≤x≤2

Find composition f(|x|) and its domain.

Solution :

Statement of the problem : The function is defined by different rules in two intervals.

The composition consists of two functions “f(x)” and “|x|”. We know that modulus is defined for all values of “x”. However, domain of “f(x)” is [-2,2]. Hence, domain of composition is intersection of two domains, which is [-2,2]. Here,

| -1 ; -2≤|x|≤0 and R f(|x|) = || |x|-1 ; 0≤|x|≤2 and R

The interval “-2≤[|x|≤0” can be interpreted in parts. The left part is “|x|≥-2”, which is always true. The right part “|x|≤0” is meaningless, which yields no solution for “x”. Therefore, upper interval of the function is not a valid interval. On the other hand, interval “0≤|x|≤2” has two parts. The left part “|x|≥0” is true for all values of “x”. The right part is “|x|≤2”. This expands to :

- 2 x 2

The intersection of “-2≤x≤2” and “R” is “-2≤x≤2”. Hence, composition is :

Questions & Answers

the definition for anatomy and physiology
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What is a cell
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Mohammed
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Muhammad
is the branch of biology that deals with the study of microorganisms.
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Mercy Reply
studies of microbes
Louisiaste
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Ziyad Reply
How bacteria create energy to survive?
Muhamad Reply
Bacteria doesn't produce energy they are dependent upon their substrate in case of lack of nutrients they are able to make spores which helps them to sustain in harsh environments
_Adnan
But not all bacteria make spores, l mean Eukaryotic cells have Mitochondria which acts as powerhouse for them, since bacteria don't have it, what is the substitution for it?
Muhamad
they make spores
Louisiaste
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the significance of food webs for disease transmission
Abreham
food webs brings about an infection as an individual depends on number of diseased foods or carriers dully.
Mark
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Esinniobiwa Reply
Assimilatory nitrate reduction is a process that occurs in some microorganisms, such as bacteria and archaea, in which nitrate (NO3-) is reduced to nitrite (NO2-), and then further reduced to ammonia (NH3).
Elkana
This process is called assimilatory nitrate reduction because the nitrogen that is produced is incorporated in the cells of microorganisms where it can be used in the synthesis of amino acids and other nitrogen products
Elkana
Examples of thermophilic organisms
Shu Reply
Give Examples of thermophilic organisms
Shu
advantages of normal Flora to the host
Micheal Reply
Prevent foreign microbes to the host
Abubakar
they provide healthier benefits to their hosts
ayesha
They are friends to host only when Host immune system is strong and become enemies when the host immune system is weakened . very bad relationship!
Mark
what is cell
faisal Reply
cell is the smallest unit of life
Fauziya
cell is the smallest unit of life
Akanni
ok
Innocent
cell is the structural and functional unit of life
Hasan
is the fundamental units of Life
Musa
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Micheal Reply
There are nothing like emergency disease but there are some common medical emergency which can occur simultaneously like Bleeding,heart attack,Breathing difficulties,severe pain heart stock.Hope you will get my point .Have a nice day ❣️
_Adnan
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Innocent
I think infection prevention and control is the avoidance of all things we do that gives out break of infections and promotion of health practices that promote life
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en français
Adama
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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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