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.