Question

How do you compute $\left( {f \circ g} \right)$ and $\left( {g \circ f} \right)$ if $g(x) = {x^2} - 8$ , $f(x) = {\left( { - x + 1} \right)^{\dfrac{1}{2}}}$ ?

Answer 1

Hint: Here we will compute the given function by using composition of a function. Let there be two functions $f$ and $g$. Let us note that the co domain of $f$ and the domain of $g$ are the same. Let us cut off $g$ and paste it on the function $f$.
Let $f:X \to Y$ and $g :Y \to Z$ be two functions. Then the function $h :X \to Z$ defined as $h(x) = g\left( {f(x)} \right)$ for every $x \in X$ is called the composition of $f$ with $g$. It is denoted by $g \circ f$ (Read this as $f$ composite with $g$ ).

Complete step-by-step solution:
If we are given two functions, we can create another function by composing one function into the other. The steps required to perform this operation are similar to when any function is solved for any given value. Such functions are called composite functions.
Let us understand what we need to do for $\left( {f \circ g} \right)(x)$ and $\left( {g \circ f} \right)(x)$. Before moving ahead let us understand evaluating a function.
We can see for evaluating we just plug in place of $x$ ,
$f\left( {g(x)} \right)$ is same as $\left( {f \circ g} \right)(x)$
$\left( {f \circ g} \right)(x) = f\left( {g(x)} \right)$-------(1)
$\left( {g \circ f} \right)(x) = g\left( {f(x)} \right)$-------(2)
Let us use the same thing on our problem,
$f(x) = {\left( { - x + 1} \right)^{\dfrac{1}{2}}}$ and $g(x) = {x^2} - 8$
Using (1) we get,
$\left( {f \circ g} \right)(x) = {\left( { - g(x) + 1} \right)^{\dfrac{1}{2}}}$
Now we substitute the value of $g(x) = {x^2} - 8$ in above equation we get,
$ \Rightarrow {\left( { - ({x^2} - 8) + 1} \right)^{\dfrac{1}{2}}}$
$ \Rightarrow {\left( { - {x^2} + 8 + 1} \right)^{\dfrac{1}{2}}}$
$\left( {f \circ g} \right)(x) = {\left( { - {x^2} + 9} \right)^{\dfrac{1}{2}}}$
Now the other composition $\left( {g \circ f} \right)(x)$ ,
$\left( {g \circ f} \right)(x) = {\left( {f(x)} \right)^2} - 8$
$ \Rightarrow {\left( {{{( - x + 1)}^{\dfrac{1}{2}}}} \right)^2} - 8$
Simplified the above term we get,
$ \Rightarrow - x + 1 - 8$
$ \Rightarrow - x - 7$
$\left( {g \circ f} \right)(x) = - \left( {x + 7} \right)$

Therefore $\left( {f \circ g} \right)(x) = {\left( { - g(x) + 1} \right)^{\dfrac{1}{2}}}$ and $\left( {g \circ f} \right)(x) = - \left( {x + 7} \right)$.

Note: Combining two functions by substituting one function’s formula in place of each $x$ in the other function’s formula. The composition of functions $f$ and $g$ is written $f \circ g$. And it read “ $f$ composed with $g$ “. Sometimes forget where each of the functions is defined before composing functions, which can lead to non-existing results. We also sometimes forget that composition is not a commutative operation that is $f \circ g \ne g \circ f$.