Answer:
[tex](f/g)(x)=\frac{x^4-x^3-7x^2+9x-2}{x-1} =x^3-7x+2\,\,\,for\,\,x\neq 1[/tex]
[tex](f/g)(2)=-4[/tex]
Step-by-step explanation:
[tex](f/g)(x)=\frac{x^4-x^3-7x^2+9x-2}{x-1} =x^3-7x+2\,\,\,for\,\,x\neq 1[/tex] and undefined for x = 1.
Notice that (x-1) is in fact a factor of f(x), so the quotient of the two functions introduces a "hole" for the new function at x = 1.
f(2) can be found by simply evaluating the expression for x = 2:
[tex](f/g)(2)=2^3-7(2)+2=-4[/tex]