# Let f(x) = 5x + 12 how do you find f^-1(x)?

Aug 28, 2016

See explanation for the answer ${f}^{- 1} \left(x\right) = \frac{x - 12}{5}$.

#### Explanation:

Disambiguation:

If y = f(x), then $x = {f}^{- 1} y$. If the function is bijective for $x \in \left(a , b\right)$,

then there is $1 - 1$ correspondence between x and y.. The

graphs of both $y = f \left(x\right)$ and the inverse $x = {f}^{- 1} \left(y\right)$ are identical,

in the interval.

The equation $y = {f}^{- 1} \left(x\right)$ is obtained by swapping x and y, in the

inverse relation $x = {f}^{- 1} \left(y\right)$.

The graph of $y = {f}^{- 1} \left(x\right)$ on the same graph sheet will be the

graph of y = f(x) rotated through a right angle, in the clockwise

,$y = f \left(x\right) = 5 x + 12$.. Solving for x,
$x = {f}^{- 1} \left(y\right) = \frac{y - 12}{5}$. Swapping x and y,
$y = {f}^{- 1} \left(x\right) = \frac{x - 12}{5}$