Combination of Functions

Quote:

Originally Posted by Macleef

$= \sqrt{x^2}$

How would you simplify the answer?

$\sqrt{x^2} = |x|$

Quote:

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.

Quote:

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.

Quote:

Originally Posted by Macleef

For $g \circ g$
$<br /> = g (\frac {x}{x - 1})$

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

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

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

You can write this more simply as $x;~x \neq 1$

-Dan