F(x)=3x-7 g(x)=x^2+3x-4...........

f(x)=3x-7
g(x)=x^2+3x-4


find (f+g)(x)
find (f-g)(x)
find (f x g)(x)
find (f/g)(x)
find (g/f)(x)
find (fog)(x)
find (gof)(x)

Both functions (f and g) can be defined in R.
(f+g)(x) = f(x) + g(x) = 3x - 7 + x^2 + 3x - 4 = x^2 + 6x - 11
(f-g)(x) = f(x) - g(x) = 3x - 7 - x^2 - 3x + 4 = -x^2 - 3
(fg)(x) = f(x)g(x) = (3x-7)(x^2+3x-4) = 3x^3 + 9x^2 -12x -7x^2 -21x +28 = 3x^3 + 2x^2 -33x + 28
(f/g)(x) = f(x)/g(x) = (3x-7) / (x^2+3x-4)
CAUTION!! Solve x^2+3x-4=0 <=> x=1 or x=-4. The function f/g is not defined for x=1 or x=-4
(g/f)(x) = g(x)/f(x) = (x^2+3x-4) / (3x-7)
CAUTION!! The function g/f is not defined for x=7/3
(fog)(x) = f(g(x)) = 3(x^2+3x-4) - 7 = 3x^2 + 9x - 12 - 7 = 3x^2 + 9x - 19
(gof)(x) = g(f(x)) = (3x-7)^2 + 3(3x-7) - 4 = 9x^2 -42x+49 + 9x - 21 = 9x^2 - 33x + 28

Hope I helped :)

All you have to do, if memory serves I believe is:

add the 2 equations together (ie: (3x-7) + (x^2+3x-4) = [x^2+6x-11, I think])
subtract the two equations
multiply the two
divide the first by the second.
divide the second by the first
substitute the second equation where it says "x", into the first equation (ie: 3(x^2+3x-4)-7)
substitute the the first equation into where it says "x" into the second equation

f(x) = 3x - 7
g(x) = x^2 + 3x - 4

(f + g)(x) = x^2 + 6x - 11
(f - g)(x) = -x^2 - 3
(f x g)(x) = 3x^3 + 2x^2 - 33x + 28
(f / g)(x) = (3x - 7) / (x + 4)(x - 1)
(g / f)(x) = (x + 4)(x - 1) / (3x - 7)
(f o g)(x) = 3x(x + 3) - 19
(g o f)(x)
= (3x - 7)^2 + 3(3x - 7) - 4
= 9x^2 - 42x + 49 + 9x - 21 - 4
= 9x^2 - 33x + 24
= 3(3x^2 - 11x + 8)
= 3(x - 1)(3x - 8)