What is the inverse function of #h(x)= 3-(x+4)/(x-7)#?

Question

Chloe Alexander · Accepted Answer

#h^-1(y) = 7 - 11/(y-2)# with restriction #y != 2# First re-factor the equation defining #h(x)# to have just one occurrence of #x#, then move the operations away from the term in #x# to leave it isolated. #y = h(x) = 3 - (x+4)/(x-7)# #=3-((x-7)+11)/(x-7)# #=3-cancel(x-7)/cancel(x-7)-11/(x-7)# #=3 - 1 - 11/(x-7)# #=2-11/(x-7)# Subtract #2# from both ends to get: #y - 2 = -11/(x-7)# Multiply both sides by #(x-7)# to get: #(x-7)(y - 2) = -11# Divide both sides by #(y-2)# to get: #x -7 = -11/(y - 2)# Add #7# to both sides to get: #x = 7-11/(y-2)# This defines #x# in terms of #y = h(x)#, so: #h^-1(y) = 7 - 11/(y-2)# with restriction #y != 2#

Serenity Johnson · Answer

undefined To find the inverse function of $ h(x) = 3 - \frac{x+4}{x-7} $, you first need to switch the roles of $ x $ and $ y $, then solve for $ y $.

$$ x = 3 - \frac{y+4}{y-7} $$

Then, solve for $ y $ in terms of $ x $.

$$ x(y - 7) = 3(y + 4) - (y + 4) $$

$$ xy - 7x = 3y + 12 - y - 4 $$

$$ xy - 7x = 2y + 8 $$

$$ xy - 2y = 7x + 8 $$

$$ y(x - 2) = 7x + 8 $$

$$ y = \frac{7x + 8}{x - 2} $$

So, the inverse function of $ h(x) $ is $ h^{-1}(x) = \frac{7x + 8}{x - 2} $.