Question

Find ﻿f+gf+gf+g﻿, ﻿f−gf-gf−g﻿, ﻿fgfgfg﻿, and ﻿fg\frac{f}{g}gf​﻿. Determine the domain for each function.﻿f(x)=xf\left(x\right)=\sqrt{x}f(x)=x​﻿, ﻿g(x)=x−5g\left(x\right)=x-5g(x)=x−5

Accepted Answer

To find $$ (f+g)(x) $$, add the functions: $$ f(x) + g(x) = \sqrt{x} + (x - 5) . $$Simplify the expression to get $$ \sqrt{x} + x - 5 . $$The domain of $$ f(x) = \sqrt{x}  is$$ $$ x \geq 0 $$, and the domain of $$ g(x) = x - 5  is $$all real numbers. Therefore, the domain of $$ f+g  is$$ $$ x \geq 0 . To $$find $$ (f-g)(x) $$, subtract the functions: $$ f(x) - g(x) = \sqrt{x} - (x - 5) . $$Simplify the expression to get $$ \sqrt{x} - x + 5 . $$The domain is the same as $$ f+g $$, which is $$ x \geq 0 . To $$find $$ (fg)(x) $$, multiply the functions: $$ f(x) \cdot g(x) = \sqrt{x} \cdot (x - 5) . $$This simplifies to $$ x\sqrt{x} - 5\sqrt{x} . $$The domain is $$ x \geq 0 $$ because $$ \sqrt{x}  is $$only defined for non-negative $$ x . To $$find $$ \left(\frac{f}{g}ight)(x) $$, divide the functions: $$ \frac{f(x)}{g(x)} = \frac{\sqrt{x}}{x - 5} . $$The domain is $$ x \geq 0 $$ and $$ x 
eq 5 $$ because the denominator cannot be zero. Summarize the domains: $$ f+g $$ and $$ f-g $$ have domain $$ x \geq 0 $$, $$ fg $$ has domain $$ x \geq 0 $$, and $$ \frac{f}{g} $$ has domain $$ x \geq 0 $$ and $$ x 
eq 5 .$$

Find $f+g$ , $f-g$ , $fg$ , and $\frac{f}{g}$ . Determine the domain for each function.
$f\(\left\)(x\(\right\))=\(\sqrt{x}\)$ , $g\(\left\)(x\(\right\))=x-5$

Verified video answer for a similar problem:

Key Concepts

Domain of a Function

Square Root Function

Linear Function

Find f+gf+gf+g, f−gf-gf−g, fgfgfg, and fg\(\frac{f}{g}\)gf. Determine the domain for each function.f(x)=xf\(\left\)(x\(\right\))=\(\sqrt{x}\)f(x)=x, g(x)=x−5g\(\left\)(x\(\right\))=x-5g(x)=x−5