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Question: Solve the following. \(f\left( x \right) = \dfrac{1}{{3 - x}}\), \(g\left( x \right) = fof\), \(h...

Solve the following.
f(x)=13xf\left( x \right) = \dfrac{1}{{3 - x}}, g(x)=fofg\left( x \right) = fof, h(x)=fofofh\left( x \right) = fofof, then 1f(x)g(x)h(x)=?\dfrac{1}{{f\left( x \right)g\left( x \right)h\left( x \right)}} = ?

Explanation

Solution

Here we are given three functions f(x)f\left( x \right), g(x)g\left( x \right) and h(x)h\left( x \right).
f(x)=13xf\left( x \right) = \dfrac{1}{{3 - x}}
g(x)=fofg\left( x \right) = fof
h(x)=fofofh\left( x \right) = fofof
Now, fof means that g(x) is the function of f, that means we can find g(x) by substituting f(x)=13xf\left( x \right) = \dfrac{1}{{3 - x}} in f(x)f\left( x \right). Now, fofof means that h(x) is a function of fof and we have found the value for fof as g(x). So, we can find the value of h(x) by substituting g(x) in f(x)=13xf\left( x \right) = \dfrac{1}{{3 - x}}.

Complete step by step solution:
In this question, we are given three functions f(x)f\left( x \right), g(x)g\left( x \right) and h(x)h\left( x \right) and we are given the value for f(x)f\left( x \right).
Given data is:
f(x)=13xf\left( x \right) = \dfrac{1}{{3 - x}}
g(x)=fofg\left( x \right) = fof
h(x)=fofofh\left( x \right) = fofof
And we need to find,
1f(x)g(x)h(x)=?\dfrac{1}{{f\left( x \right)g\left( x \right)h\left( x \right)}} = ?
So, first of all, g(x) is fof. That means g(x) is a function of f. That means when we substitute the function ff in the variable xx in function ff, we get foffof. Therefore, we get
g(x)=fof g(x)=f(f(x))  \Rightarrow g\left( x \right) = fof \\\ \Rightarrow g\left( x \right) = f\left( {f\left( x \right)} \right) \\\
Now, we need to put f(x)=13xf\left( x \right) = \dfrac{1}{{3 - x}} in f(x)f(x)
g(x)=1313x g(x)=13(3x)13x g(x)=(3x)93x1 g(x)=(3x)83x  \Rightarrow g\left( x \right) = \dfrac{1}{{3 - \dfrac{1}{{3 - x}}}} \\\ \Rightarrow g\left( x \right) = \dfrac{1}{{\dfrac{{3\left( {3 - x} \right) - 1}}{{3 - x}}}} \\\ \Rightarrow g\left( x \right) = \dfrac{{\left( {3 - x} \right)}}{{9 - 3x - 1}} \\\ \Rightarrow g\left( x \right) = \dfrac{{\left( {3 - x} \right)}}{{8 - 3x}} \\\
Hence, we have found the value for g(x) and now we need to find the value for h(x).
Now, h(x) is fofoffofof that means hh is a function of foffof and we have found the value of foffof as g(x)g (x). Therefore, we get
h(x)=fofof h(x)=f(fof) h(x)=f(g(x))  \Rightarrow h\left( x \right) = fofof \\\ \Rightarrow h\left( x \right) = f\left( {fof} \right) \\\ \Rightarrow h\left( x \right) = f\left( {g\left( x \right)} \right) \\\
Now, we need to put g(x)=(3x)83xg\left( x \right) = \dfrac{{\left( {3 - x} \right)}}{{8 - 3x}} in f(x)=13xf\left( x \right) = \dfrac{1}{{3 - x}}. Therefore, we get
h(x)=13x h(x)=133x83x h(x)=83x3(83x)3+x h(x)=83x249x3+x h(x)=83x218x  \Rightarrow h\left( x \right) = \dfrac{1}{{3 - x}} \\\ \Rightarrow h\left( x \right) = \dfrac{1}{{3 - \dfrac{{3 - x}}{{8 - 3x}}}} \\\ \Rightarrow h\left( x \right) = \dfrac{{8 - 3x}}{{3\left( {8 - 3x} \right) - 3 + x}} \\\ \Rightarrow h\left( x \right) = \dfrac{{8 - 3x}}{{24 - 9x - 3 + x}} \\\ \Rightarrow h\left( x \right) = \dfrac{{8 - 3x}}{{21 - 8x}} \\\
Therefore, we now have all the values we need. Therefore, substituting these values, we get
1f(x)g(x)h(x)=1(13x)(3x83x)(83x218x) 1f(x)g(x)h(x)=11(218x) 1f(x)g(x)h(x)=218x  \Rightarrow \dfrac{1}{{f\left( x \right)g\left( x \right)h\left( x \right)}} = \dfrac{1}{{\left( {\dfrac{1}{{3 - x}}} \right)\left( {\dfrac{{3 - x}}{{8 - 3x}}} \right)\left( {\dfrac{{8 - 3x}}{{21 - 8x}}} \right)}} \\\ \Rightarrow \dfrac{1}{{f\left( x \right)g\left( x \right)h\left( x \right)}} = \dfrac{1}{{\dfrac{1}{{\left( {21 - 8x} \right)}}}} \\\ \Rightarrow \dfrac{1}{{f\left( x \right)g\left( x \right)h\left( x \right)}} = 21 - 8x \\\
Hence, we have found the value of 1f(x)g(x)h(x)=218x\dfrac{1}{{f\left( x \right)g\left( x \right)h\left( x \right)}} = 21 - 8x.

Note:
Properties of composite functions are
Associative Property: If f, g and h are given three functions, then they are said to be associative if
f(gh)=(fg)hf \circ \left( {g \circ h} \right) = \left( {f \circ g} \right) \circ h
Commutative property: If f and g are given two functions, then they are said to be commutative if
gf=fgg \circ f = f \circ g