Question

How do you find the composite function $\left( fogoh \right)\left( x \right)$ for $f\left( x \right)=\dfrac{x-2}{2x+1}$, $g\left( x \right)=3x+1$ and $h\left( x \right)={{x}^{2}}$?

Answer 1

Hint: We start solving the problem by using the fact that $\left( fogoh \right)\left( x \right)=f\left( g\left( h\left( x \right) \right) \right)$. We then make the necessary calculations and then substitute $g\left( x \right)=3x+1$ in the obtained result to proceed through the problem. We then make the necessary calculations and then substitute $h\left( x \right)={{x}^{2}}$ to proceed further through the problem. We then make the necessary calculations to get the required composite function $\left( fogoh \right)\left( x \right)$.

Complete step by step answer:
According to the problem, we are given $f\left( x \right)=\dfrac{x-2}{2x+1}$, $g\left( x \right)=3x+1$ and $h\left( x \right)={{x}^{2}}$. We need to find the value of $\left( fogoh \right)\left( x \right)$.
We know that $\left( fogoh \right)\left( x \right)=f\left( g\left( h\left( x \right) \right) \right)$.
So, we have \[\left( fogoh \right)\left( x \right)=\dfrac{g\left( h\left( x \right) \right)-2}{2\left( g\left( h\left( x \right) \right) \right)+1}\] ---(1).
Let us use $g\left( x \right)=3x+1$ in equation (1).
\[\Rightarrow \left( fogoh \right)\left( x \right)=\dfrac{3\left( h\left( x \right) \right)+1-2}{2\left( 3\left( h\left( x \right) \right)+1 \right)+1}\].
\[\Rightarrow \left( fogoh \right)\left( x \right)=\dfrac{3\left( h\left( x \right) \right)-1}{6\left( h\left( x \right) \right)+2+1}\].
\[\Rightarrow \left( fogoh \right)\left( x \right)=\dfrac{3\left( h\left( x \right) \right)-1}{6\left( h\left( x \right) \right)+3}\] ---(2).
Now, let us substitute $h\left( x \right)={{x}^{2}}$ in equation (2) to find the composite function $\left( fogoh \right)\left( x \right)$.
\[\Rightarrow \left( fogoh \right)\left( x \right)=\dfrac{3\left( {{x}^{2}} \right)-1}{6\left( {{x}^{2}} \right)+3}\].
\[\Rightarrow \left( fogoh \right)\left( x \right)=\dfrac{3{{x}^{2}}-1}{6{{x}^{2}}+3}\].
So, we have found the composite function $\left( fogoh \right)\left( x \right)$ as \[\dfrac{3{{x}^{2}}-1}{6{{x}^{2}}+3}\].
$\therefore $ The required composite function $\left( fogoh \right)\left( x \right)$ is \[\dfrac{3{{x}^{2}}-1}{6{{x}^{2}}+3}\].

Note:
We should not confuse $\left( fogoh \right)\left( x \right)$ with $h\left( g\left( f\left( x \right) \right) \right)$ instead of $f\left( g\left( h\left( x \right) \right) \right)$, which is the common mistake done by students. We should perform each step carefully in order to avoid confusion and calculation mistakes while solving this problem. We can also solve the given problem as shown below:
We know that $\left( fogoh \right)\left( x \right)=f\left( g\left( h\left( x \right) \right) \right)$ ---(3).
Now, let us substitute $h\left( x \right)={{x}^{2}}$ in equation (4).
So, we have $\left( fogoh \right)\left( x \right)=f\left( g\left( {{x}^{2}} \right) \right)$ ---(4).
Now, let us find the value of $g\left( {{x}^{2}} \right)$.
So, we have $g\left( {{x}^{2}} \right)=3\left( {{x}^{2}} \right)+1$.
$\Rightarrow g\left( {{x}^{2}} \right)=3{{x}^{2}}+1$ ---(5).
Let us substitute equation (5) in equation (4).
$\Rightarrow \left( fogoh \right)\left( x \right)=f\left( 3{{x}^{2}}+1 \right)$.
Now, let us find $f\left( 3{{x}^{2}}+1 \right)$ to get the required composite function.
\[\Rightarrow \left( fogoh \right)\left( x \right)=\dfrac{\left( 3{{x}^{2}}+1 \right)-2}{2\left( 3{{x}^{2}}+1 \right)+1}\].
\[\Rightarrow \left( fogoh \right)\left( x \right)=\dfrac{3{{x}^{2}}+1-2}{6{{x}^{2}}+2+1}\].
\[\Rightarrow \left( fogoh \right)\left( x \right)=\dfrac{3{{x}^{2}}-1}{6{{x}^{2}}+3}\]
So, the composite function $\left( fogoh \right)\left( x \right)$ is \[\dfrac{3{{x}^{2}}-1}{6{{x}^{2}}+3}\].