Find f(g(x)) - g(f(x)) where f(x) = (x) / (x + 1) and g(x) = (x + 1) / (x)?

Answer 1

Ah hi Jose ; wow i must confess i got headache when i looked at the other answers .

well lets get started;
f(x) = (x) / (x + 1)
g(x) = (x + 1) / (x) = (2x + 1)/ x

Part 1;
f(g(x)) = ( (x + 1) / (x) ) / (((x + 1) / (x)) + 1=
( (x + 1) / (x) ) / ((2x +1) / x) =
x ( x +1 ) / x * ( 2x +1 ) { you can remove 'x' in this fraction
so
f(g(x)) = (x+1) / (2x +1)

Part 2 ;
g(f(x)) = (( (x) / (x + 1))+1) / (x / x +1) =
((2x +1) /( x +1) ) / ((x) / (x+1))
(x+1)(2x +1) / (x+1)(x) { you can remove ' x+1' in this fraction }
so
g(f(x)) = (2x+1) / (x)

Part 3;
f(g(x)) - g(f(x)) = (x+1) / (2x +1) - (2x+1) / (x) =
[ x(x+1) - (2x+1)(2x+1)] / (2x+1)(x) =
x(x+1) - (2x+1)^2 / (2x+1)(x) =
(-3x^2 -3x -1) / (+2x^2 + x)

Good Luck Josy .

Answer 2

Use substitution:

f(g(x))-g(f(x)) =

f((y+1)/y) - g(y/(y+1)) =

Notice that I changed the letters. Actually the letter does not matter, as long as it is the same quantity that is represented. For example, take x+2 = 5. The solution is x = 3. But suppose I had used y for my unknown. Then it would have been y+2 = 5, and the answer is y = 3. Get the same number in either case. I changed the letters to avoid confusion, since both functions are expressed in terms of x.

f((y+1)/y) - g(y/(y+1)) = [(y+1)/y]/[((y+1)/y)+1] - [(y+1)/y + 1]/((y+1)/y)

I substituted y/(y+1) for x in the first expression and (y+1)/y for x in the second one.

The rest is algebra, and involves complex fractions. You should be able to handle it from here.

Answer 3

f(x) = x/(x + 1)
g(x) = (x + 1)/x

f(g(x)) = f((x + 1)/x)
f((x + 1)/x) = ((x + 1)/x)/(((x + 1)/x) + 1)
f((x + 1)/x) = ((x + 1)/x)/(((x + 1) + x)/x)
f((x + 1)/x) = ((x + 1)/x)/((x + 1 + x)/x)
f((x + 1)/x) = ((x + 1)/x)/((2x + 1)/x)
f((x + 1)/x) = ((x + 1)/x) * (x/(2x + 1))
f((x + 1)/x) = (x(x + 1))/(x(2x + 1))
f((x + 1)/x) = (x + 1)/(2x + 1)

g(f(x)) = g(x/(x + 1))
g(x/(x + 1)) = ((x/(x + 1))) + 1)/(x/(x + 1))
g(x/(x + 1)) = ((x + (x + 1))/(x + 1))/(x/(x + 1))
g(x/(x + 1)) = ((x + x + 1)/(x + 1))/(x/(x + 1))
g(x/(x + 1)) = ((2x + 1)/(x + 1))/(x/(x + 1))
g(x/(x + 1)) = ((2x + 1)/(x + 1)) * ((x + 1)/x)
g(x/(x + 1)) = ((2x + 1)(x + 1))/(x(x + 1))
g(x/(x + 1)) = (2x + 1)/x

so

f(g(x)) - g(f(x)) = ((x + 1)/(2x + 1)) - ((2x + 1)/x)
f(g(x)) - g(f(x)) = (x(x + 1) - ((2x + 1)(2x + 1)))/(x(2x + 1))
f(g(x)) - g(f(x)) = (x^2 + x - (4x^2 + 2x + 2x + 1))/(x(2x + 1))
f(g(x)) - g(f(x)) = (x^2 + x - (4x^2 + 4x + 1))/(x(2x + 1))
f(g(x)) - g(f(x)) = (x^2 + x - 4x^2 - 4x - 1)/(x(2x + 1))
f(g(x)) - g(f(x)) = (-3x^2 - 3x - 1)/(x(2x + 1))
f(g(x)) - g(f(x)) = (-3x^2 - 3x - 1)/(2x^2 + x)

Answer 4

f(g(x)) = f((x + 1) / (x))
Now, wherever you see x in f(x), replace it by (x + 1) / (x)) to find
f(g(x)).
f(g(x)) = [(x + 1) / x] / [((x + 1) / (x)) +1]
= [(x + 1) / x] / [(2x + 1)/x]
= [(x + 1) / x]*[x/(2x + 1)]
= (x + 1)/(2x + 1)

g(f(x)) = g(x / (x + 1))
Wherever you see x in g(x), replace it by x / (x + 1) to find g(f(x)).
g(f(x)) = [x / (x + 1) + 1] / [x / (x + 1)]
= [(2x+1)/(x+1)] / [x / (x + 1)]
= [(2x+1)/(x+1)]*[(x + 1)/ x]
= (2x + 1)/x

f(g(x)) - g(f(x)) = (x + 1)/(2x + 1) - (2x + 1)/x
= [x(x + 1) - (2x + 1)^2] / [x(2x + 1)]
= (x^2 + x - 4x^2 - 4x - 1) / (2x^2 + x)
= (-3x^2 -3x - 1) / (2x^2 + x)