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Q37E

Expert-verified
Found in: Page 115

### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

# Question: Find ${\mathbit{f}}{\mathbf{}}{\mathbf{+}}{\mathbf{}}{\mathbit{g}}$ and ${\mathbit{f}}{\mathbit{g}}$ for the functions f and g given in exercise 36.

The function $f+g$ is $\left(f+g\right)\left(x\right)={x}^{2}+x+3$

$fg$ is $\left(fg\right)\left(x\right)={x}^{3}+2{x}^{2}+x+2$

See the step by step solution

## Step 1:

Composition of f and g : ${\mathbf{\left(}}{\mathbit{f}}{\mathbf{\circ }}{\mathbit{g}}{\mathbf{\right)}}{\mathbf{\left(}}{\mathbit{a}}{\mathbf{\right)}}{\mathbf{=}}{\mathbit{f}}{\mathbf{\left(}}{\mathbit{g}}{\mathbf{\left(}}{\mathbit{a}}{\mathbf{\right)}}{\mathbf{\right)}}$

## Step 2:

Given:

$g:\mathrm{ℝ}\to \mathrm{ℝ}\text{and f}:\mathrm{ℝ}\to \mathrm{ℝ}\phantom{\rule{0ex}{0ex}}f\left(x\right)={x}^{2}+1\phantom{\rule{0ex}{0ex}}f\left(x\right)=x+2$

Since f and g are both from$\mathrm{ℝ}\to \mathrm{ℝ}$, $f\circ g\text{and}g\circ f$are also functions from$\mathrm{ℝ}\to \mathrm{ℝ}$

Use the definition of composition:

$\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)=\left({x}^{2}+1\right)+\left(x+2\right)={x}^{2}+x+3\phantom{\rule{0ex}{0ex}}\left(fg\right)\left(x\right)=f\left(x\right).g\left(x\right)=\left({x}^{2}+1\right)\left(x+2\right)={x}^{3}+2{x}^{2}+x+2$

Hence, the solution is $f+g$ is $\left(f+g\right)\left(x\right)={x}^{2}+x+3$

$fg$ is $\left(fg\right)\left(x\right)={x}^{3}+2{x}^{2}+x+2$