Question

Find (fog)(x) when f(x)=x^2 + 6x + 5 and g(x)= 1/x+1

Answer 1

For this case we have the following functions:

`f (x) = x ^ 2 + 6x + 5\\g (x) = \frac {1} {x + 1}`

We must find
`(f_ {0} g) (x).` By definition of composition of functions we have to:

`(f_ {0} g) (x) = f (g (x))`

So, we have:

`f (g (x)) = (\frac {1} {x + 1}) ^ 2 + 6 (\frac {1} {x + 1}) + 5\\f (g (x)) = \frac {1} {(x + 1) ^ 2} + \frac {6} {x + 1} +5\\f (g (x)) = \frac {1 + 6 (x + 1) +5 (x + 1) ^ 2} {(x + 1) ^ 2}\\f (g (x)) = \frac {1 + 6 (x + 1) +5 (x + 1) ^ 2} {(x + 1) ^ 2}\\f (g (x)) = \frac {1 + 6x + 6 + 5 (x ^ 2 + 2x + 1)} {(x + 1) ^ 2}\\f (g (x)) = \frac {1 + 6x + 6 + 5x ^ 2 + 10x + 5} {(x + 1) ^ 2}\\f (g (x)) = \frac {5x ^ 2 + 16x + 12} {(x + 1) ^ 2}\\f (g (x)) = \frac {(5x + 6) (x + 2)} {(x + 1) ^ 2}`

Answer:

`f (g (x)) = \frac {(5x + 6) (x + 2)} {(x + 1) ^ 2}`