1. ## Combination of Functions

Find the inverse of each function.

g)

$\displaystyle y = \frac {x + 5}{x -5}$

Where did I go wrong?

$\displaystyle x = \frac {y + 5}{y - 5}$

$\displaystyle x(y - 5) = y + 5$

$\displaystyle x(y - 5) -5 = y$

$\displaystyle xy - 5x - 5 = y$

$\displaystyle -5x - 5 = y - xy$

$\displaystyle -5x - 5 = y(1 - x)$

$\displaystyle \frac {-5x - 5}{1 - x} = y$

$\displaystyle \frac {-5(x + 1)}{1 - x} = y$

Even if I move it to the right side. . .I get a different answer. . .

y = $\displaystyle \frac {5(x - 1)}{x - 1}$

$\displaystyle \frac {5(x + 1)}{x - 1}$

h)

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

$\displaystyle x(y^2) = 1$

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

$\displaystyle y = \sqrt{\frac {1}{x}}$

2. Originally Posted by Macleef
Find the inverse of each function.

g)

$\displaystyle y = \frac {x + 5}{x -5}$

Where did I go wrong?

$\displaystyle x = \frac {y + 5}{y - 5}$

$\displaystyle x(y - 5) = y + 5$

$\displaystyle x(y - 5) -5 = y$

$\displaystyle xy - 5x - 5 = y$

$\displaystyle -5x - 5 = y - xy$

$\displaystyle -5x - 5 = y(1 - x)$

$\displaystyle \frac {-5x - 5}{1 - x} = y$

$\displaystyle \frac {-5(x + 1)}{1 - x} = y$

Even if I move it to the right side. . .I get a different answer. . .

y = $\displaystyle \frac {5(x - 1)}{x - 1}$

$\displaystyle \frac {5(x + 1)}{x - 1}$
You have:

$\displaystyle \frac {-5(x + 1)}{1 - x} = y$

or:

$\displaystyle y=\frac {-5(x + 1)}{1 - x}=\frac{5(x+1)}{(x-1)}$

(that is multiply top and bottom of the middle term side by $\displaystyle -1$ to get the right most term, which is the desired answer)

3. Originally Posted by Macleef
h)

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

$\displaystyle x(y^2) = 1$

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

$\displaystyle y = \sqrt{\frac {1}{x}}$
Subject to the restriction that $\displaystyle x > 0$ yes except that you have reversed the meaning of $\displaystyle x$ and $\displaystyle y$.

$\displaystyle y(x^2) = 1$

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

$\displaystyle x = \sqrt{\frac {1}{y}}$

when $\displaystyle y > 0$

RonL