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How do you find the inverse of $f (x) = \frac{1}{3} x + 2$ $f(x)= 1/3x + 2$ and is it a function?

1 Answer

Jul 18, 2016

$g (x) = 3 x - 6$ $color(blue)(g(x)=3x-6)$ is a function
(which is the inverse of $f (x) = \frac{1}{3} x + 2$ $f(x)=1/3x+2$ )

Explanation:

If $g (x)$ $g(x)$ is the inverse of $f (x)$ $f(x)$
then by definition
$XXX g (f (x)) = x$ $color(white)("XXX")g(f(x))=x$
and
$XXX f (g (x)) = x$ $color(white)("XXX")f(g(x))=x$

Since $f (x) = \frac{1}{3} x + 2$ $f(color(red)(x))=1/3color(red)(x)+2$
then
$XXX f (g (x)) = \frac{1}{3} g (x) + 2$ $color(white)("XXX")f(color(red)(g(x)))=1/3color(red)(g(x))+2$

But from our original definition we know that
$XXX f (g (x)) = x$ $color(white)("XXX")f(g(x))=x$

Therefore
$XXX f (g (x)) = \frac{1}{3} g (x) + 2 = x$ $color(white)("XXX")f(g(x))=1/3g(x)+2=x$

$XXX \Rightarrow \frac{1}{3} g (x) = x - 2$ $color(white)("XXX")rArr1/3g(x)=x-2$

$XXX \Rightarrow g (x) = 3 x - 6$ $color(white)("XXX")rArrg(x)=3x-6$

Since any value of $x$ $x$ gives a single value for $g (x)$ $g(x)$
it follows that $g (x)$ $g(x)$ is a function.

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