# Combination of Functions

• Jan 14th 2008, 04:01 PM
Macleef
Combination of Functions
In each of the following cases find:
$\displaystyle f \circ g$
$\displaystyle g \circ f$
$\displaystyle f \circ f$
$\displaystyle g \circ g$

c)

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

$\displaystyle g(x) = x^2$

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

$\displaystyle = f(x^2)$

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

How would you simplify the answer?

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

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

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

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

For $\displaystyle g \circ f$

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

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

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

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

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

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

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

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

$\displaystyle = \frac {x^2 - x}{x - 1}$
• Jan 14th 2008, 04:10 PM
topsquark
Quote:

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

How would you simplify the answer?

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

Quote:

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

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

How would you simplify the answer?

$\displaystyle (\sqrt{x})^2 = x$ <-- 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)
$\displaystyle f(x) = \frac {1}{x + 1}$

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

For $\displaystyle g \circ f$

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

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

Quote:

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

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

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

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

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

-Dan