# If f: R - {3} -> R - {4} , and the function f(x)=(mx+4)/(nx-9) is one to one and onto, what is f(1/3) ?

Jan 23, 2017

$f \left(\frac{1}{3}\right) = - 1$

#### Explanation:

Since the domain of $f$ is $\mathbb{R} - \left\{3\right\}$, $x$ cannot take the value of $3$.

The point where $n x - 9 = 0$ is not in the domain, therefore $3 n - 9 = 0$ is not in the domain.

$3 n - 9 = 0$

$\iff$

$3 n = 9$

$\iff$

$n = 3$

So, by this point we have $f \left(x\right) = \frac{m x + 4}{3 x - 9}$

If $f$ is one-to-one and onto, then $f$ is bijective. If $f$ is bijective, then it has an inverse.

$\exists {f}^{- 1} : \mathbb{R} - \left\{4\right\} \to \mathbb{R} - \left\{3\right\}$

First we need to find it

Let $y = f \left(x\right)$ then

$y = \frac{m x + 4}{3 x - 9}$

$\iff$ multiply both sides by $3 x - 9$

$\left(3 x - 9\right) y = \left(m x + 4\right)$

$\iff$ distribute $y$

$3 x y - 9 y = m x + 4$

$\iff$ subtract $4$ from both sides

$3 x y - 9 y - 4 = m x$

$\iff$ subtract $m x$ from both sides

$3 x y - 9 y - 4 - m x = 0$

$\iff$ group like terms

$3 x y - m x - 9 y - 4 = 0$

$\iff$ factor $x$ from $3 x y - m x$

$x \left(3 y - m\right) - 9 y - 4 = 0$

$\iff$ subtract $x \left(3 y - m\right)$ from both sides

$- 9 y - 4 = - x \left(3 y - m\right)$

$\iff$ divide both sides by $\left(3 y - m\right)$

$- \frac{9 y - 4}{3 y - m} = - x$

$\iff$ multiply both sides by $- 1$

$\frac{9 y - 4}{3 y - m} = x$

and since $\left({f}^{- 1} \circ f\right) \left(x\right) = {f}^{- 1} \left(f \left(x\right)\right) = x$

$\implies {f}^{- 1} \left(x\right) = \frac{9 x - 4}{3 x - m}$

and since the domain of ${f}^{- 1}$ is $\mathbb{R} - \left\{4\right\}$, we have that

$12 - m = 0$

$\iff$

$12 = m$

Then this leads us to $f \left(x\right) = \frac{12 x + 4}{3 x - 9}$

Then $f \left(\frac{1}{3}\right) = \frac{12 \left(\frac{1}{3}\right) + 4}{3 \left(\frac{1}{3}\right) - 9} = \frac{4 + 4}{1 - 9} = \frac{8}{- 8} = - 1$