Question 1196503

Determine the value of k so that the function {{{f(x)=(x-6)/(2x+k)}}} will be its own inverse.


inverse function {{{f(x) = (x - 6)/(2 x + k)}}} is :


recall that {{{f(x) =y}}}


{{{y= (x - 6)/(2 x + k)}}}.........swap variables

{{{x= (y - 6)/(2y + k)}}}.........solve for {{{y}}}

{{{x(2y + k)= (y - 6)}}}

 {{{ 2xy + kx= y - 6}}}

{{{2xy -y = -kx - 6}}}

{{{(2x -1)y = -kx - 6}}}

{{{y =(-kx- 6)/(2x-1)}}}

so, inverse is

{{{f(x)}}}’ = {{{(-k x - 6)/(2 x - 1)}}}


now equal {{{f(x)}}} and {{{f(x)}}}’

{{{(x-6)/(2x+k)= (-k x - 6)/(2 x - 1)}}}

{{{(x-6)(2 x - 1)=(2x+k) (-k x - 6)}}}

{{{(x-6)(2 x - 1)=(2x+k) (-k x - 6)}}}

{{{2x^2 - 13x + 6= - 2kx^2 - 12x-k^2*x- 6k }}}

{{{2x^2 - 13x + 6= - 2kx^2 - (12+k^2)x- 6k}}} .......quadratics will be equal if and only if  corresponding coefficients are equal

{{{highlight(2)x^2 - 13x + 6= highlight(-2k)x^2 - (12+k^2)x- 6k}}}


then

=>{{{2=-2k }}}

=>{{{highlight(k=-1)}}}


and your function is:


 {{{f(x)=(x-6)/(2x-1)}}} and will be its own inverse