Find invarient point(s) Algebraically

• Nov 13th 2008, 03:56 PM
Raj
Find invarient point(s) Algebraically
Given $\displaystyle f(x) = 3x -6$, algebraically determine $\displaystyle y = f^-1(x)$ and the invarient point(s).

$\displaystyle y = 3x - 6$
$\displaystyle x = 3y - 6$
$\displaystyle x + 6 = 3y$
$\displaystyle y = \frac{x}{3} + 2$

That is as far as I got. I know the answer is (3,3), but how do i show this.
• Nov 13th 2008, 04:16 PM
mr fantastic
Quote:

Originally Posted by Raj
Given $\displaystyle f(x) = 3x -6$, algebraically determine $\displaystyle y = f^-1(x)$ and the invarient point(s).

$\displaystyle y = 3x - 6$
$\displaystyle x = 3y - 6$
$\displaystyle x + 6 = 3y$
$\displaystyle y = \frac{x}{3} + 2$

That is as far as I got. I know the answer is (3,3), but how do i show this.

Solve $\displaystyle f(x) = f^{-1}(x)$, that is, $\displaystyle 3x - 6 = \frac{x}{3} + 2$.
• Nov 13th 2008, 04:22 PM
Raj
Quote:

Originally Posted by mr fantastic
Solve $\displaystyle f(x) = f^{-1}(x)$, that is, $\displaystyle 3x - 6 = \frac{x}{3} + 2$.

Ok $\displaystyle x = 3$

What about the y-value? Do I just sub that into the second or first equation?
• Nov 13th 2008, 04:36 PM
mr fantastic
Quote:

Originally Posted by Raj
Ok $\displaystyle x = 3$

What about the y-value? Do I just sub that into the second or first equation?

Do you think it matters ..... ? Substitute into both. What do you find? What other line must this point lie on? Does this surprise you?