Let f and g be real functions, defined by `f(x)=sqrt(x+2)andg(x)=sqrt(4-x^(2))`. Find
(i) (f+g)(x) (ii) (f-g)(x) (iii)(fg)(x) (iv) `(ff)(x)`
(v) (gg)(x) (vi) `((f)/(g))(x)`.

Explanation

Clearly, `f(x)=sqrt(x+2)` is defined for all `x""inR` such that `x+2ge0,i.e.,xge-2`.
`:."dom "(f)=[-2,oo)`.
Again, `g(x)=sqrt(4-x^(2))` is defined for all `x""inR` such that `4-x^(2)ge0`.
But, `4-x^(2)ge0impliesx^(2)-4le0implies(x+2)(x-2)leimpliesx""in[-2,2]`.
`:."dom "(g)=[-2,2]`
`:."dom "(f)nn"dom "(g)=[-2,oo)nn[-2,2]=[-2,2]`.
(i) `(f+g):[-2,2]toR` is given by
`(f+g)(x)=f(x)+g(x)=sqrt(x+2)-sqrt(4-x^(2))`
(ii) `(f-g):[-2,2]toR` is given by
`(f-g)(x)=f(x)=g(x)=sqrt(x+2)-sqrt(4-x^(2))` ltbgt (iii) `(fg):[-2,2]toR` is given by
`(fg)(x)=f(x).g(x)=(sqrt(x+2))(sqrt(2-x^(2)))`
`=sqrt((x+2)^(2)(2-x))=(x+2)sqrt((2-x))`.
(iv) `(ff):[-2,2]toR` is given by
`(ff)(x)=f(x)_.f(x)=(sqrt(x+2))(sqrt(x+2))=(x+2)`.
(v) `(gg):[-2,2]toR` is given by `(gg)(x)=g(x).g(x)=(sqrt(4-x^(2)))(sqrt(4-x^(2)))=(4-x^(2))`.
(vi) `{x:g(x)=0}={x:4-x^(2)=0]={x:(2-x)(2+x)=0}={-2,2}`
`"dom "((f)/(g))="dom "(f)nn"dom "{g}-{x:g(x)=0}`
`=[-2,2]-[-2,2]=(-2,2)`.
`:.(f)/(g):(-2,2)toR` is given by
`((f)/(g))(x)=(f(x))/(g(x))=(sqrt(x+2))/(sqrt(4-x^(2)))=(sqrt(2+x))/((sqrt(2+x))(sqrt(2-x)))=(1)/((sqrt(2-x)))`.