Question 1159812
 
{{{matrix(2,1,"",f(x)=(2x^2)/(x^2-1) )}}}

 {{{matrix(2,1,"",g(x)=e^(x^2-4))}}}


Solve for {{{x}}} the equation {{{g(x)=16}}}.


{{{matrix(2,1,"",16=e^(x^2-4))}}}.......take natural log of both sides

{{{matrix(2,1,"",ln(16)=ln(e^(x^2-4)))}}}

{{{matrix(2,1,"",ln(2^4)=ln(e^(x^2-4)))}}}

{{{matrix(2,1,"",4ln(2)=(x^2-4) ln(e))}}}

{{{matrix(2,1,"",4ln(2)/ln(e)=(x^2-4)) }}}.........{{{ln(e)=1}}}

{{{x^2-4=4ln(2) }}}

{{{x^2=4ln(2) +4}}}

{{{x^2=4(ln(2) +1)}}}

{{{x=sqrt(4(ln(2) +1))}}}

{{{x=2sqrt(ln(2) +1)}}}


Find {{{f(g(x))}}}. Simplify your answer.


 {{{matrix(2,1,"",f(g(x))=f(e^(x^2-4)))}}}


{{{matrix(2,1,"",f(g(x))=(2(e^(x^2-4))^2)/((e^(x^2-4))^2-1))}}}


{{{matrix(2,1,"",f(g(x))=(2(e^(2(x^2-4))))/(e^(2(x^2-4))-1))}}}


{{{matrix(2,1,"",f(g(x))=(2e^(2x^2-8))/(e^(2x^2-8)-1))}}}


{{{matrix(2,1,"",f(g(x))=(2e^(2x^2-8))/(((e^(x^2) - e^4)(e^(x^2) + e^4))/e^8))}}}


{{{matrix(2,1,"",f(g(x))=(2e^(2x^2))/((e^(x^2) - e^4) (e^(x^2) + e^4)))}}}



Describe the domains of {{{f(x)}}} and {{{g(x)}}}.


the domain of{{{ f(x)}}}


{{{matrix(2,1,"",f(x)=(2x^2)/(x^2-1) )}}}

domain will be all real numbers except values of {{{x}}} that make denominator equal to zero, so exclude those

{{{x^2-1=0 }}} ->{{{x^2=1 }}}->{{{x=sqrt(1 )}}}=>{{{x=1}}} and {{{x=-1}}}


domain is:

{ {{{x}}} element {{{R}}} : {{{x<>-1}}} and {{{x<>1}}} }



the domain of{{{ g(x)}}}

 {{{matrix(2,1,"",g(x)=e^(x^2-4))}}}

{{{R}}} (all real numbers)