1. ## Combination of Functions

In each of the following cases find:
$f \circ g$
$g \circ f$
$f \circ f$
$g \circ g$

c)

$f(x) = \sqrt{x}$

$g(x) = x^2$

For $f \circ g$
$= f(g(x))$

$= f(x^2)$

$= \sqrt{x^2}$

How would you simplify the answer?

For $g \circ f$
$= g(f(x))$

$= (\sqrt {x})^2$

How would you simplify the answer?

Could you please check the following to see if I did them correctly and show me where I messed up?

d)
$f(x) = \frac {1}{x + 1}$

$g(x) = \frac {x}{x -1}$

For $g \circ f$

$= g (\frac {x}{x + 1})$

$= \frac {(\frac{x}{x + 1})}{( \frac{x}{x + 1} - 1)}$

$= \frac {(\frac {x}{x + 1})}{(\frac{1}{x + 1})}$

$= \frac {x}{x + 1} (x + 1)$

$= \frac {x^2 + x}{x + 1}$

For $g \circ g$
$
= g (\frac {x}{x - 1})$

$= \frac {(\frac {x}{x - 1})}{(\frac {x}{x -1} - 1)}$

$= (\frac {x}{x - 1}) (x - 1)$

$= \frac {x^2 - x}{x - 1}$

2. Originally Posted by Macleef
$= \sqrt{x^2}$

How would you simplify the answer?
$\sqrt{x^2} = |x|$

Originally Posted by Macleef
For $g \circ f$
$= g(f(x))$

$= (\sqrt {x})^2$

How would you simplify the answer?
$(\sqrt{x})^2 = x$ <-- Since x can't be negative anyway, we don't need to worry about negative answers.

Originally Posted by Macleef
Could you please check the following to see if I did them correctly and show me where I messed up?

d)
$f(x) = \frac {1}{x + 1}$

$g(x) = \frac {x}{x -1}$

For $g \circ f$

$= g (\frac {x}{x + 1})$
$f(x) = \frac{1}{x + 1}$
You are off by a factor of x.

Originally Posted by Macleef
For $g \circ g$
$
= g (\frac {x}{x - 1})$

$= \frac {(\frac {x}{x - 1})}{(\frac {x}{x -1} - 1)}$

$= (\frac {x}{x - 1}) (x - 1)$

$= \frac {x^2 - x}{x - 1}$
You can write this more simply as $x;~x \neq 1$

-Dan