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Q37E

Expert-verifiedFound in: Page 115

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

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

Answer:

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

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

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

Given:

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

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

Use the definition of composition:

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

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

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

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