`(f/g)(x)=x^2/(2x+1),x!=-1/2` why?

If f(x) =x^(2) -2,g (x) =2x+ 1 then f^(@) g (x)

If f(x)=x^2+(1)/(x^2), g(x)=x^4+(1)/(x^4) and a+(1)/(a)=3 , then the respectively values of f(a) and g(a) are

If f(x)=1+x^2 and [g(x)]=1+x^2-2x^3+x^4, g(2)=2 , then

g(x)= (x-(1/2))/x-1 , f(x)=2x-1 , then fog(x) is

If f(x) =x^(2) +1, g(x) = x^(2) - 5x+6 , find f+g, f-g, f/g .

Find f0g and g0f : f(x)=x^2+2 , g(x)=1-1/(1-x)

If f(x)=2 x^(2)+x+1 and g(x)=3 x+1 then f o g(2)

If f^(1)(x) =g(x) and g^(1)(x) =-f (x) for all x and f(2) = 4 =f^(1) (2) then f^(2) (4) +g^(2)(4) is

If g(x)=x^(2)+x-2 and (1)/(2)g(f(x))=2x^(2)-5x+2, then f(x) is