Question

Need help on these, thank you!

Answer 1

`5x+2 , 2x^2+5x-2,(x^2+5x)/(2-x^2)`

Explanation:

We are given f(x) and g(x)

1. (f+g)(x)

(f+g)(x) = f(x) + g(x)

=
`x^2+5x+2-x^2`

=
`5x+2`

Domain : All real numbers as it there exists a value of (f+g)(x) f every x .

2. (f-g)(x)

(f-g)(x) = f(x)-g(x)

=
`x^2+5x-2+x^2`

=
`2x^2+5x-2`

Domain : All real numbers as it there exists a value of (f-g)(x) f every x .

Part 3 .

`((f)/(g))(x)\\((f)/(g))(x) = (f(x))/(g(x))\\=(x^2+5x)/(2-x^2)`

Domain : In this case we see that the function is not defined for values of x for which the denominator becomes 0 or less than zero . Hence only those values of x are defined for which

`2-x^2>0`

or
`2>x^2`

Hence taking square roots on both sides and solving inequality we get.

`-√(2) <x<√(2)`