How would you show that #f (x) = 7x +3# and #f^-1(x) = (x +3 )/ 7# are inverses of each other?

Question

How would you show that #f (x) = 7x  +3# and #f^-1(x) = (x +3 )/ 7# are inverses of each other?

Christian Antunez · Accepted Answer

undefined To show that $ f(x) = 7x + 3 $ and $ f^{-1}(x) = \frac{x + 3}{7} $ are inverses of each other, we need to verify that $ f(f^{-1}(x)) = x $ and $ f^{-1}(f(x)) = x $ for all $ x $ in their respective domains.

Let's first find $ f(f^{-1}(x)) $:
$$ f(f^{-1}(x)) = f\left(\frac{x + 3}{7}\right) = 7\left(\frac{x + 3}{7}\right) + 3 = x + 3 $$

Now, let's find $ f^{-1}(f(x)) $:
$$ f^{-1}(f(x)) = f^{-1}(7x + 3) = \frac{7x + 3 + 3}{7} = x $$

Since both $ f(f^{-1}(x)) $ and $ f^{-1}(f(x)) $ simplify to $ x $, we can conclude that $ f(x) $ and $ f^{-1}(x) $ are inverses of each other.

Christian Antunez · Answer

Verify that #f^(-1)(x) = (x-3)/7# not #(x+3)/7#

These functions are not inverses of one another.

The correct formula for #f^(-1)(x)# can be written:

#f^(-1)(x) = (x-3)/7#

In fact the variable name used does not matter, so I will write:

#f^(-1)(y) = (y-3)/7#

Then substituting #y=f(x)# we find:

#f^(-1)(f(x)) = (f(x)-3)/7 = ((7x+3)-3)/7 = (7x)/7 = x#

Substituting #x=f^(-1)(y)# we find:

#f(f^(-1)(y)) = 7f^(-1)(y)+3 = 7((y-3)/7)+3 = (y-3)+3 = y#