Question 1131456


{{{f(x)=x^2-x-12}}} and {{{g(x)= x-4}}}

 Perform each function operation and find the domain. 

{{{f(x)-2g(x)=x^2-x-12-2(x-4)}}}

{{{f(x)-2g(x)=x^2-x-12-2x+8}}}

{{{f(x)-2g(x)=x^2-3x-4}}}


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

({{{-infinity}}},{{{infinity}}})



{{{f(x)*g(x)=(x^2-x-12)(x-4)}}}

{{{f(x)*g(x)=x^3-x^2-12x-4x^2+4x+48}}}

{{{f(x)*g(x)=x^3-5x^2-8x+48}}}


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

({{{-infinity}}},{{{infinity}}})



{{{f(x)/g(x)=(x^2-x-12)/(x-4)}}}

{{{f(x)/g(x)=(x^2-4x+3x-12)/(x-4)}}}

{{{f(x)/g(x)=((x^2-4x)+(3x-12))/(x-4)}}}

{{{f(x)/g(x)=(x(x-4)+3(x-4))/(x-4)}}}

{{{f(x)/g(x)=((x+3)cross((x-4)))/cross((x-4))}}}-> we can cancel {{{x-4}}}, but it will be used for a domain

{{{f(x)/g(x)=x+3}}}


domain:

{ {{{x}}} element {{{R}}} : {{{x<>4}}} }




{{{g(x)/f(x)=(x-4)/(x^2-x-12)}}}


{{{g(x)/f(x)=cross((x-4))/((x+3)cross((x-4)))}}}

{{{g(x)/f(x)=1/(x+3)}}}


domain:

{ {{{x}}} element {{{R}}} : {{{x<>4}}} and {{{x<>-3}}} }