Answer
The value of x from the functions $f\left( x \right)=2x-5$ , $g\left( x \right)={{x}^{2}}-3x+8$ and $\left( f\circ g \right)\left( x \right)=7$ is $\left\{ 1,2 \right\}$.
Work Step by Step
The composition of f with g can be defined as the function $\left( f\circ g \right)\left( x \right)$:
$f\left( g\left( x \right) \right)$ where $f\left( x \right)=2x-5$ and $g\left( x \right)={{x}^{2}}-3x+8$.
Consider $\left( f\circ g \right)\left( x \right)=f\left( g\left( x \right) \right)$.
Now substitute the value of $g\left( x \right)$ in $f\left( g\left( x \right) \right)$ such that:
$f\left( g\left( x \right) \right)=f\left( {{x}^{2}}-3x+8 \right)$
Calculate for the function $f$:
$\begin{align}
& f\left( {{x}^{2}}-3x+8 \right)=2\left( {{x}^{2}}-3x+8 \right)-5 \\
& =2{{x}^{2}}-6x+16-5 \\
& =2{{x}^{2}}-6x+11
\end{align}$
Thus,
$\left( f\circ g \right)\left( x \right)=2{{x}^{2}}-6x+11$.
Also,
$\left( f\circ g \right)\left( x \right)=7$.
Hence,
$2{{x}^{2}}-6x+11=7$
Now, solve the above equation for the value of x:
Move all the zero terms on one side of the equation and take the non-zero terms on the other side of the equation.
$\begin{align}
& 2{{x}^{2}}-6x+11=7 \\
& 2{{x}^{2}}-6x+4=0
\end{align}$
Make the factors of the above equation:
$\begin{align}
& 2{{x}^{2}}-6x+4=0 \\
& 2{{x}^{2}}-2x-4x+4=0 \\
& 2x\left( x-1 \right)-4\left( x-1 \right)=0 \\
& \left( 2x-4 \right)\left( x-1 \right)=0
\end{align}$
Put each of the factors equal to zero to get the value of x:
$\begin{align}
& \left( 2x-4 \right)=0 \\
& 2x=4 \\
& x=2
\end{align}$
Or
$\begin{align}
& x-1=0 \\
& x=1
\end{align}$
Hence, the value of x from the functions $f\left( x \right)=2x-5$ , $g\left( x \right)={{x}^{2}}-3x+8$ and $\left( f\circ g \right)\left( x \right)=7$ is $\left\{ 1,2 \right\}$.
