Question #95d2b

1 Answer
Nov 29, 2017

#f^-1(x)=+-sqrt((x+5)/7)# (In the explanation, we will not use #f^-1(x)# to make things simpler.

Explanation:

#g(x)# is inverse of #f(x)# if #g(f(x))=x#. If this is true, this is also true: #f(g(x))=x#.

Let's call our inverse function #g(x)#
We know that #g(f(x))=x# and that #f(g(x))=x#.

Out of the two, we will use #f(g(x))=x#.
Whatever #g(x)# is, we know that when it is plugged into #f(x)#, it will give us #x#. To make matters simpler, we will call #g(x)# as #y#.

Therefore, we know that #7y^2-5=x# Isolate #y#.
#7y^2-5=x#
#7y^2=x+5#
#y^2=(x+5)/7#
#y=+-sqrt((x+5)/7)#

We see that when these are plugged into #7x^2-5#, it gives us #x#.