How do you verify if $f (x) = 7 x; g (x) = \frac{1}{7} x$ $f(x)=7x; g(x)=1/7x$ are inverse functions?

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Answer 1 · 2017-04-14T22:33:52

See below.

We swap the places of $x$ $x$ and $y$ $y$ (in this case, $f (x)$ $f(x)$ ).

$y = 7 x$ $y=7x$

$x = 7 y$ $x=7y$

$\frac{1}{7} x = y$ $1/7x=y$ , which is $g (x)$ $g(x)$ , so the functions are inverses of each other.

Answer 2 · 2017-04-15T15:51:14

Here's another way of going at it. Apply the identity $f (f^{- 1} (x)) = x$ $f(f^-1(x)) = x$ .

$f (f^{- 1} (x)) = f (g (x)) = 7 (\frac{1}{7} x) = x \sqrt$ $f(f^-1(x)) = f(g(x)) = 7(1/7x) = x color(green)(√)$

Check the other way:

$f^{- 1} (f (x)) = g (f (x)) = \frac{1}{7} (7 x) = x \sqrt$ $f^-1(f(x)) = g(f(x)) = 1/7(7x) = x color(green)(√)$

Hence, $f (x)$ $f(x)$ and $g (x)$ $g(x)$ are indeed inverses of each other.

Hopefully this helps!

How do you verify if f(x)=7x;g(x)=17xf(x)=7x; g(x)=1/7x are inverse functions?