Question

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

Answer 1

Hint: Here we are given three functions $f\left( x \right)$, $g\left( x \right)$ and $h\left( x \right)$.
$f\left( x \right) = \dfrac{1}{{3 - x}}$
$g\left( x \right) = fof$
$h\left( x \right) = fofof$
Now, fof means that g(x) is the function of f, that means we can find g(x) by substituting $f\left( x \right) = \dfrac{1}{{3 - x}}$ in $f\left( x \right)$. 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 $f\left( x \right) = \dfrac{1}{{3 - x}}$.

Complete step by step solution:
In this question, we are given three functions $f\left( x \right)$, $g\left( x \right)$ and $h\left( x \right)$ and we are given the value for $f\left( x \right)$.
Given data is:
$f\left( x \right) = \dfrac{1}{{3 - x}}$
$g\left( x \right) = fof$
$h\left( x \right) = fofof$
And we need to find,
$\dfrac{1}{{f\left( x \right)g\left( x \right)h\left( x \right)}} = ?$
So, first of all, g(x) is fof. That means g(x) is a function of f. That means when we substitute the function $f$ in the variable $x$ in function $f$, we get $fof$. Therefore, we get
$
   \Rightarrow g\left( x \right) = fof \\
   \Rightarrow g\left( x \right) = f\left( {f\left( x \right)} \right) \\
 $
Now, we need to put $f\left( x \right) = \dfrac{1}{{3 - x}}$ in $f(x)$
$
   \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 $fofof$ that means $h$ is a function of $fof$ and we have found the value of $fof$ as $g (x)$. Therefore, we get
$
   \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 $g\left( x \right) = \dfrac{{\left( {3 - x} \right)}}{{8 - 3x}}$ in $f\left( x \right) = \dfrac{1}{{3 - x}}$. Therefore, we get
$
   \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
$
   \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 $\dfrac{1}{{f\left( x \right)g\left( x \right)h\left( x \right)}} = 21 - 8x$.

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
$f \circ \left( {g \circ h} \right) = \left( {f \circ g} \right) \circ h$
Commutative property: If f and g are given two functions, then they are said to be commutative if
$g \circ f = f \circ g$