Answer:
[tex]5x+2 , 2x^2+5x-2,\frac{x^2+5x}{2-x^2}[/tex]
Step-by-step explanation:
We are given f(x) and g(x)
1. (f+g)(x)
(f+g)(x) = f(x) + g(x)
= [tex]x^2+5x+2-x^2[/tex]
= [tex]5x+2[/tex]
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)
= [tex]x^2+5x-2+x^2[/tex]
=[tex]2x^2+5x-2[/tex]
Domain : All real numbers as it there exists a value of (f-g)(x) f every x .
Part 3 .
[tex](\frac{f}{g})(x)\\(\frac{f}{g})(x) = \frac{f(x)}{g(x)}\\=\frac{x^2+5x}{2-x^2}[/tex]
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
[tex]2-x^2>0[/tex]
or [tex]2>x^2[/tex]
Hence taking square roots on both sides and solving inequality we get.
[tex]-\sqrt{2} <x<\sqrt{2}[/tex]
