# Let f(x) = 2x - 6, how do you solve f^-1(x) when x = 2?

Jun 8, 2017

${f}^{-} 1 \left(2\right) = 4$

#### Explanation:

Let

$y = 2 x - 6$

To get ${f}^{-} 1 \left(x\right)$, solve for $x$ in terms of $y$:

$y = 2 x - 6$

$y + 6 = 2 x$

$\frac{1}{2} y + 3 = x$ or $x = \frac{1}{2} y + 3$

Which means ${f}^{-} 1 \left(x\right) = \frac{1}{2} x + 3$

Plugging in $x = 2$ gives

${f}^{-} 1 \left(2\right) = \frac{1}{2} \left(2\right) + 3$

$= 1 + 3 = 4$