Question

How do you find (f o g) (x) and its domain, (g o f) (x) and its domain, (f o g) (-2) and (g o f) (-2) of the following problem \[f(x) = {x^2} - 1\] , \[g(x) = x + 1\] ?

Answer 1

Hint: We can write \[(f{\text{ }}0{\text{ }}g)(x)\] as \[f(g(x))\] . Similarly we can write \[(g{\text{ }}0{\text{ }}f)(x)\] as \[g(f(x))\] . Here we have a composition of two functions. The composition is an operation where two functions say ‘f’ and ‘g’ generate a new function say ‘h’ in such a way that \[h(x) = g(f(x))\] . It means here function g is applied to the function of ‘x’.

Complete step-by-step answer:
As we know that, \[(f{\text{ }}0{\text{ }}g)(x) = f(g(x))\] and \[(g{\text{ }}0{\text{ }}f)(x) = g(f(x))\] .
Given, \[f(x) = {x^2} - 1\] and \[g(x) = x + 1\]
Now,
 \[ \Rightarrow (f{\text{ }}0{\text{ }}g)(x) = f(g(x))\]
We have \[f(x) = {x^2} - 1\] in this we need to put \[x = g(x)\] . Then we have,
 \[ = {[g(x)] ^2} - 1\]
But we have \[g(x) = x + 1\] . Squaring this on both sides we have \[{\left( {g(x)} \right)^2} = {\left( {x + 1} \right)^2}\] .
 \[ = {\left( {x + 1} \right)^2} - 1\]
We know \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] , applying this we have,
 \[ = {x^2} + 1 + 2x - 1\]
 \[ = {x^2} + 2x\]
Thus we have \[ \Rightarrow (f{\text{ }}0{\text{ }}g)(x) = {x^2} + 2x{\text{ }} - - - (1)\] .
Since this is defined for all real values of ‘x’, the domain of \[(f{\text{ }}0{\text{ }}g)(x)\] is all real numbers.
We also need to find \[(f{\text{ }}0{\text{ }}g)( - 2)\] .
Put \[x = - 2\] in the equation (1) we have
 \[ \Rightarrow (f{\text{ }}0{\text{ }}g)( - 2) = {( - 2)^2} + 2( - 2)\]
 \[ = 4 - 4\]
 \[ = 0\] .
Similarly, now take \[(g{\text{ }}0{\text{ }}f)(x) = g(f(x))\]
We have \[g(x) = x + 1\] in this we need to put \[x = f(x)\] . Then we have,
 \[ = f(x) + 1\]
But we have \[f(x) = {x^2} - 1\] , then substituting in the above we get,
 \[ = {x^2} - 1 + 1\]
 \[ = {x^2}\] .
Thus we have, \[ \Rightarrow (g{\text{ }}0{\text{ }}f)(x) = {x^2}{\text{ }} - - - (2)\] .
Again, this is defined for all real values of ‘x’ so the domain of \[(g{\text{ }}0{\text{ }}f)(x)\] is all real value.
Now we need \[(g{\text{ }}0{\text{ }}f)( - 2)\]
So put \[x = - 2\] in the equation (2) we get,
 \[ \Rightarrow (g{\text{ }}0{\text{ }}f)( - 2) = {( - 2)^2}\]
 \[ = 4\] .

Note: Follow the same procedure for any composition problem. As we can see in above problem it fails commutative property because \[(f{\text{ }}0{\text{ }}g)(x) \ne (g{\text{ }}0{\text{ }}f)(x)\] . If ‘f’ and ‘g’ are one-one functions then the composition function \[(f{\text{ }}0{\text{ }}g)(x)\] is also one-one function. The function composition of two onto function is always onto.