Question 1128469


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

a)
Find {{{(f+g)(2)=f(2)+g(2)=2^2+2*2-3=4+4-3=8-3=5}}}

b)
Find the domain of {{{(f/g)(x)=f(x)/g(x)=x^2/(2x-3)}}}

denominator cannot be equal to zero, so exclude value of {{{x}}} which makes it equal to zero

{{{2x-3=0}}}=>{{{x=3/2}}}

so, the domain: { {{{x}}} elememt {{{R}}}: {{{x<>3/2}}}

c)

Find the range of {{{f+g}}}

{{{f+g=x^2+2x-3}}}-> it's upward parabola,has minimum ;so,  write it in vertex form 

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

{{{f+g=(x^2+2x+1^2)-1^2-3}}}

{{{f+g=(x+1)^2-4}}} => vertex (minimum) is at ({{{1}}},{{{-4}}})

so, the range is :

{ {{{y}}} element {{{R}}} : {{{y>=-4}}} }