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

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Answer 1 · 2017-06-08T22:45:17

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

Let

$y = 2 x - 6$ $y=2x-6$

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

$y = 2 x - 6$ $y=2x-6$

$y + 6 = 2 x$ $y+6=2x$

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

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

Plugging in $x = 2$ $x=2$ gives

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

$= 1 + 3 = 4$ $=1+3=4$

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