Answer
The values of x from the functions $f\left( x \right)=1-2x$ , $g\left( x \right)=3{{x}^{2}}+x-1$ and $\left( f\circ g \right)\left( x \right)=-5$ are $\left\{ \frac{-4}{3},1 \right\}$.
Work Step by Step
The composition of f with g can be defined as the function $\left( f\circ g \right)$ , and therefore $\left( f\circ g \right)\left( x \right)$ can be written as $f\left( g\left( x \right) \right)$.Where $f\left( x \right)=1-2x$ and $g\left( x \right)=3{{x}^{2}}+x-1$
Consider the equation below:
$\left( f\circ g \right)\left( x \right)=f\left( g\left( x \right) \right)$
Now, substitute the value of $g\left( x \right)$ in $f\left( g\left( x \right) \right)$ such that
$f\left( g\left( x \right) \right)=f\left( 3{{x}^{2}}+x-1 \right)$
Solve it for the function $f$:
$\begin{align}
& f\left( 3{{x}^{2}}+x-1 \right)=1-2\left( 3{{x}^{2}}+x-1 \right) \\
& =1-6{{x}^{2}}-2x+2 \\
& =-6{{x}^{2}}-2x+3
\end{align}$
Thus
$\left( f\circ g \right)\left( x \right)=-6{{x}^{2}}-2x+3$
The value of $\left( f\circ g \right)\left( x \right)=-5$
Therefore,
$-6{{x}^{2}}-2x+3=-5$
Now, solve the above equation for the value of x:
Move all the non-zero terms to one side of the equation and solve:
$\begin{align}
& -6{{x}^{2}}-2x+3=-5 \\
& 6{{x}^{2}}+2x-8=0
\end{align}$
Make the factors of the above equation:
$\begin{align}
& 6{{x}^{2}}+2x-8=0 \\
& 6{{x}^{2}}-6x+8x-8=0 \\
& 6x\left( x-1 \right)+8\left( x-1 \right)=0 \\
& \left( 6x+8 \right)\left( x-1 \right)=0
\end{align}$
Put each of the factors equal to zero to find the value of x:
$\begin{align}
& \left( 6x+8 \right)=0 \\
& 6x=-8 \\
& x=\frac{-8}{6} \\
& =\frac{-4}{3}
\end{align}$
Or
$\begin{align}
& x-1=0 \\
& x=1
\end{align}$
Hence, the values of x from the functions $f\left( x \right)=1-2x$ , $g\left( x \right)=3{{x}^{2}}+x-1$ and $\left( f\circ g \right)\left( x \right)=-5$ are $\left\{ \frac{-4}{3},1 \right\}$.
