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)=6−1xf\left(x\right)=6-\frac{1}{x}f(x)=6−x1​﻿, ﻿g(x)=1xg\left(x\right)=\frac{1}{x}g(x)=x1​

Accepted Answer

To find $$ (f+g)(x) $$, add the functions: $$ f(x) = 6 - \frac{1}{x} $$ and $$ g(x) = \frac{1}{x} . $$Combine like terms: $$ (f+g)(x) = 6 - \frac{1}{x} + \frac{1}{x} = 6 . $$The domain of $$ f+g  is $$all real numbers except $$ x = 0 $$ because division by zero is undefined. To find $$ (f-g)(x) $$, subtract the functions: $$ f(x) = 6 - \frac{1}{x} $$ and $$ g(x) = \frac{1}{x} . $$Combine like terms: $$ (f-g)(x) = 6 - \frac{1}{x} - \frac{1}{x} = 6 - \frac{2}{x} . $$The domain of $$ f-g  is $$all real numbers except $$ x = 0 . To $$find $$ (fg)(x) $$, multiply the functions: $$ f(x) = 6 - \frac{1}{x} $$ and $$ g(x) = \frac{1}{x} . $$Distribute $$ g(x) $$ into $$ f(x) $$: $$ (fg)(x) = (6 - \frac{1}{x}) \cdot \frac{1}{x} = \frac{6}{x} - \frac{1}{x^2} . $$The domain of $$ fg  is $$all real numbers except $$ x = 0 . To $$find $$ \frac{f}{g}(x) $$, divide the functions: $$ f(x) = 6 - \frac{1}{x} $$ and $$ g(x) = \frac{1}{x} . $$Perform the division: $$ \frac{f}{g}(x) = \frac{6 - \frac{1}{x}}{\frac{1}{x}} = x(6 - \frac{1}{x}) = 6x - 1 . $$The domain of $$ \frac{f}{g}  is $$all real numbers except $$ x = 0 . $$For each function, ensure the domain excludes $$ x = 0 $$ because both $$ f(x) $$ and $$ g(x) $$ involve division by $$ x $$, which is undefined at $$ x = 0 .$$

Find $f+g$ , $f-g$ , $fg$ , and $\frac{f}{g}$ . Determine the domain for each function.
$f\(\left\)(x\(\right\))=6-\(\frac{1}{x}\)$ , $g\(\left\)(x\(\right\))=\(\frac{1}{x}\)$

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Find f+gf+gf+g, f−gf-gf−g, fgfgfg, and fg\(\frac{f}{g}\)gf. Determine the domain for each function.f(x)=6−1xf\(\left\)(x\(\right\))=6-\(\frac{1}{x}\)f(x)=6−x1, g(x)=1xg\(\left\)(x\(\right\))=\(\frac{1}{x}\)g(x)=x1