## Precalculus (6th Edition) Blitzer

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 $\left\{ \frac{-4}{3},1 \right\}$.
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 $\left\{ \frac{-4}{3},1 \right\}$.