Question 1186274
a. 
{{{f(x) = 2/(3 x) - 4}}}

inverse
{{{y = 2/(3 x) - 4}}}........swap variables
{{{x = 2/(3y) - 4}}}........solve for {{{y}}}
{{{x+4 = 2/(3y) }}}
{{{3y = 2/(x+4)}}}
{{{y  = 2/(3(x+4))}}}
{{{f}}}'{{{(x)  = 2/(3(x+4))}}}

Domain
{ {{{x}}} element {{{R}}} : {{{x<>-4}}} }
Range
{ {{{f(x)}}} element {{{R}}} : {{{f(x)<>0}}} }


{{{ graph( 600, 600, -10, 10, -10, 10, 2/(3x)-4,2/(3(x+4))) }}}


b. 
{{{f(x) = 2x^2 -1}}}

inverse
{{{y = 2x^2 -1}}}
{{{x = 2y^2 -1}}}
{{{x+1 = 2y^2 }}}
{{{(x+1) /2=y^2 }}}
{{{y=sqrt((x+1) /2) }}}

{{{f}}}'{{{(x)}}}=  ± {{{sqrt(x + 1)/sqrt(2)}}}

not a function


{{{ graph( 600, 600, -10, 10, -10, 10, 2x^2 -1,sqrt(x + 1)/sqrt(2), -sqrt(x + 1)/sqrt(2)) }}}


c. 
{{{f(x) = 1/(x-1)}}}
inverse
{{{y = 1/(x-1)}}}
{{{x= 1/(y-1)}}}
{{{(y-1)= 1/x}}}
{{{y= 1/x+1}}}
{{{f}}}'{{{(x)=1/x+1}}}

Domain
{ {{{x }}} element {{{R}}} : {{{x<>0}}} }
Range 
{ {{{f(x)}}} element {{{R }}}: {{{f(x)<=1}}} } 


{{{ graph( 600, 600, -10, 10, -10, 10, 1/(x-1),1/x+1) }}}