Question 1140384

In order to show that {{{f(x)}}} is one-to-one, we need to show that if {{{f(x[1])=f(x[2])}}}, then {{{x[1]=x[2]}}}. 


In other words, we need to show that each {{{y}}} value ONLY has ONE unique {{{x}}} value mapping to it. 
In this case, we're assuming that {{{x[1]}}} and {{{x[2]}}} are two different values that map to the same {{{y }}}}value (but as you'll see, they are in fact the same value)


{{{f(x[1])=f(x[2])}}} Start with the given equation

{{{6/(x[1]+5)=6/(x[2]+5)}}} ...divide by {{{6}}}

{{{1/(x[1]+5)=1/(x[2]+5)}}} .......cross multiply

{{{(x[2]+5)=(x[1]+5)}}} .......cancel {{{5}}}

{{{x[2]=x[1]}}} 

So we've shown that if {{{f(x[1])=f(x[2])}}}, then {{{x[1]=x[2]}}}. This means that {{{f(x)}}} is one-to-one.

now, find a formula for the inverse:

{{{f(x) = 6/(x + 5) }}}........recall that {{{f(x)=y}}}

{{{y= 6/(x + 5) }}}.......swap {{{x}}} and {{{y}}}

{{{x= 6/(y + 5) }}}.........solve for  {{{y}}}

{{{x(y + 5) = 6}}}

{{{y + 5 = 6/x}}}

{{{y  = 6/x-5}}}

{{{f^-1(x) =  6/x-5}}}



{{{ graph( 600, 600, -10, 10, -10, 10, 6/(x+5), 6/x-5) }}}