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

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Answer 1 · 2016-08-28T07:32:25

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

Disambiguation:

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

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

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

in the interval.

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

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

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

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

sense, about the origin.

Here

, $y = f (x) = 5 x + 12$ $y = f(x) = 5x+12$ .. Solving for x,

$x = f^{- 1} (y) = \frac{y - 12}{5}$ $x = f^(-1)(y) = ( y - 12 )/5$ . Swapping x and y,

$y = f^{- 1} (x) = \frac{x - 12}{5}$ $y = f^(-1)(x) = (x-12)/5$

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