Question 1189906


 (1)

{{{x^2y+y=6}}}

{{{(x^2 + 1) y = 6}}}

{{{y = 6/(x^2 + 1)}}}

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

{{{f(x)=6/(x^2 + 1)}}}


(2)

domain:

all real numbers: {{{x}}} element of {{{R}}} 


(3) state any three integer which are not in the range of {{{f}}}

range: { {{{f}}} element {{{R}}} : {{{0 < f <= 6}}} }

any three integer which are not in the range: {{{-1}}},{{{0}}},{{{7}}}


(4) obtain a formula for{{{ f^-1}}}


{{{f(x)=6/(x^2 + 1)}}}.........{{{y =f(x)}}}

{{{y=6/(x^2 + 1)}}}......swap variables

{{{x=6/(y^2 + 1)}}}...........solve for {{{y}}}

{{{y^2 + 1=6/x}}}

{{{y^2 =6/x-1}}}

{{{y}}} = ±{{{sqrt(6/x-1)}}}

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



(5) compute {{{f^-1(5)}}}

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



(6) For what value(s) of {{{x}}} is {{{f^-1}}} undefined

if denominator equal to zero, so if {{{x=0}}}  then {{{f^-1}}} will be  undefined


(7) for what value(s) of {{{x}}} is {{{f^-1}}} zero


 {{{0}}}=± {{{sqrt(6 - x)/sqrt(x)}}}

if numerator equal to zero, so

if {{{x = 6}}} then {{{f^-1 =0}}}