Question

How do you find $\left[ {fog} \right]\left( x \right)$ and $\left[ {gof} \right]\left( x \right)$ given $f\left( x \right) = 2x - 3$ and $g\left( x \right) = {x^2} - 2x$?

Answer 1

Hint: Remember $\left[ {fog} \right]\left( x \right)$ is nothing but the composite function in other words it is the product of f and g. $\left[ {fog} \right]\left( x \right)$ can be written as $f\left( {g\left( x \right)} \right)$. First substitute the value of g and then substitute the value in f.

Complete step-by-step answer:
The objective of the problem is to find the $\left[ {fog} \right]\left( x \right)$ and $\left[ {gof} \right]\left( x \right)$.
Given functions are , $f\left( x \right) = 2x - 3$ , $g\left( x \right) = {x^2} - 2x$
Before finding the $\left[ {fog} \right]\left( x \right)$ and $\left[ {gof} \right]\left( x \right)$ let us know about the functions and composite functions.
About functions : Let A and B be two non empty sets. A function f from A to B , denoted by $f:A \to B$ is a rule that assigns each member of A an unique member of B. here A is called the domain off while B is called the co domain of f . the set of all those members of B that are assigned by the rule f to some member of A is called the range of f.
Composite function : Suppose f is a mapping from A into B , and g is mapping from B into C. Let $a \in A$ , then $f\left( a \right) \in B$. Therefore, $f\left( a \right)$ is an element in the domain of g. we can find the g image of $f\left( a \right)$ which is written as $g\left( {f\left( a \right)} \right)$. So $g\left( {f\left( a \right)} \right)$ is defined. We therefore have a rule to assign each $a \in A$, uniquely $g\left( {f\left( a \right)} \right)$ in C. thus we have a new function from A into C. this new function is called the composite function of f and g is defined by \[gof\] . The composite function is also called the product of f and g.
Now we find $\left[ {fog} \right]\left( x \right)$
We can write $\left[ {fog} \right]\left( x \right)$ as $f\left( {g\left( x \right)} \right)$.
Substitute the value $g\left( x \right) = {x^2} - 2x$ , we get
$f\left( {g\left( x \right)} \right) = f\left( {{x^2} - 2x} \right)$
Now substitute ${x^2} - 2x$ in the place of x in $f\left( x \right) = 2x - 3$, we get
$
  f\left( {g\left( x \right)} \right) = f\left( {{x^2} - 2x} \right) \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\left( {{x^2} - 2x} \right) - 3 \\
$
On simplifying above equation we get
$ \Rightarrow 2{x^2} - 4x - 3$
 Therefore , $\left[ {fog} \right]\left( x \right) = 2{x^2} - 4x - 3$
 Similarly we now find $\left[ {gof} \right]\left( x \right)$
We can write $\left[ {gof} \right]\left( x \right)$ as $g\left( {f\left( x \right)} \right)$.
Substitute the value $f\left( x \right) = 2x - 3$ , we get
$g\left( {f\left( x \right)} \right) = g\left( {2x - 3} \right)$
Now substitute $2x - 3$ in the place of x in$g\left( x \right) = {x^2} - 2x$, we get
$
  g\left( {f\left( x \right)} \right) = g\left( {2x - 3} \right) \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left[ {2x - 3} \right]^2} - 2\left[ {2x - 3} \right] \\
$
On simplifying above equation we get
$
   \Rightarrow {\left[ {2x - 3} \right]^2} - 2\left[ {2x - 3} \right] \\
   \Rightarrow {\left( {2x} \right)^2} - 2\left( {2x} \right)\left( 3 \right) + {\left( 3 \right)^2} - 2\left( {2x} \right) + 2\left( 3 \right) \\
   \Rightarrow 4{x^2} - 12x + 9 - 4x + 6 \\
   \Rightarrow 4{x^2} - 16x + 15 \\
$
Therefore , $\left[ {gof} \right]\left( x \right) = 4{x^2} - 16x + 15$.

Note: In the product function \[gof\] the co domain of f is the domain of g. The domain of \[gof\] is the domain of f. The co domain of \[gof\] and g is the same set.