## College Algebra (6th Edition)

$x=2$ and $x=1$ satisfy the given conditions
$f(x)=2x-5$ $,$ $g(x)=x^{2}-3x+8$ $,$ and $(f\circ g)(x)=7$ To find $(f\circ g)(x)$, substitute $x$ by $g(x)$ in $f(x)$ and simplify: $(f\circ g)(x)=2(x^{2}-3x+8)-5=2x^{2}-6x+16-5=...$ $...=2x^{2}-6x+11$ Set $2x^{2}-6x+11$ equal to $7$: $2x^{2}-6x+11=7$ Take $7$ to the left side of the equation: $2x^{2}-6x+11-7=0$ $2x^{2}-6x+4=0$ Take out common factor $2$ from the left side and take it to divide the right side: $2(x^{2}-3x+2)=0$ $x^{2}-3x+2=\dfrac{0}{2}$ $x^{2}-3x+2=0$ Solve by factoring: $(x-2)(x-1)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x-2=0$ $x=2$ $x-1=0$ $x=1$ $x=2$ and $x=1$ satisfy the given conditions