If $f (x) = 6 x - 4$ $f(x) = 6x-4$ , what is $f^{- 1} (x)$ $f^-1(x)$ ?

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Answer 1 · 2017-07-14T15:55:35

Here it is.....

Let $f (x) = y$ $f(x)=color(red)y$ so that it is easily replacable
Then make $x$ $x$ in terms of $y$ $y$

$y = 6 x - 4$ $color(red)(y)=6color(blue)(x)-4$

$6 x - 4 = y$ $6color(blue)(x)-4=color(red)(y)$

$6 x = y + 4$ $6color(blue)(x)=color(red)(y)+4$

$x = (y + 4) \div 6$ $color(blue)(x)=(color(red)(y)+4)-:6$
Replace the $x$ $color(blue)(x)$ by $f^{- 1} (x)$ $f^-1(x)$ and replace the $y$ $color(red)(y)$ by the $x$ $color(blue)(x)$

$f^{- 1} (x) = (x + 4) \div 6$ $f^-1(x)=(color(blue)(x)+4)-:6$

If f(x)=6x−4f(x) = 6x-4, what is f−1(x)f^-1(x)?