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Q. 1

Expert-verified
Found in: Page 119

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Finding roots of piecewise-defined functions: For each function f that follows, find all values x = c for which f(c) = 0. Check your answers by sketching a graph of f. $\begin{array}{r}f\left(x\right)=\left\{\begin{array}{rl}4-{x}^{2},& \text{if}x<0\\ x+1,& \text{if}x\ge 0\end{array}\right\\\ f\left(x\right)=\left\{\begin{array}{rl}x+1,& \text{if}x<0\\ 4-{x}^{2},& \text{if}x\ge 0\end{array}\right\\\ f\left(x\right)=\left\{\begin{array}{rl}2x-1,& \text{if}x\le 1\\ 2{x}^{2}+x-3,& \text{if}x>1\end{array}\right\\end{array}$

$f\left(x\right)=\left\{\begin{array}{rl}4-{x}^{2},& \text{if}x<0\\ x+1,& \text{if}x\ge 0\end{array}\right\$ has no roots.

$f\left(x\right)=\left\{\begin{array}{rl}x+1,& \text{if}x<0\\ 4-{x}^{2},& \text{if}x\ge 0\end{array}\right\$ we have x=-1,2

$f\left(x\right)=\left\{\begin{array}{rl}2x-1,& \text{if}x\le 1\\ 2{x}^{2}+x-3,& \text{if}x>1\end{array}\right\$we have $x=1,\frac{1}{2}$

See the step by step solution

## Step 1. Given information

We have to find the roots of the following functions :

$\begin{array}{r}f\left(x\right)=\left\{\begin{array}{rl}4-{x}^{2},& \text{if}x<0\\ x+1,& \text{if}x\ge 0\end{array}\right\\\ f\left(x\right)=\left\{\begin{array}{rl}x+1,& \text{if}x<0\\ 4-{x}^{2},& \text{if}x\ge 0\end{array}\right\\\ f\left(x\right)=\left\{\begin{array}{rl}2x-1,& \text{if}x\le 1\\ 2{x}^{2}+x-3,& \text{if}x>1\end{array}\right\\end{array}$

## Step 2. Finding roots.

$\begin{array}{r}\text{let}4-{x}^{2}=0\phantom{\rule{1em}{0ex}}\left(x<0\right)\\ \begin{array}{c}x=-2\\ \text{let}x+1=0\phantom{\rule{1em}{0ex}}\left(x\ge 0\right)\\ x=-1\end{array}\text{}\end{array}$

It can not be considered as root.

$f\left(x\right)=\left\{\begin{array}{l}x+1:x<0\\ 4-{x}^{2}:x\ge 0\end{array}\right\$

Let x+1=0

x=-1

Let $4-{x}^{2}=0\phantom{\rule{0ex}{0ex}}x=+2$

$f\left(x\right)=\left\{\begin{array}{l}2x-1:x\le 1\\ 2{x}^{2}+x-3:x>1\end{array}\right\$

Let 2x-1=0

$x=\frac{1}{2}$

Let role="math" localid="1649831777557" $2{x}^{2}+x-3=0\phantom{\rule{1em}{0ex}}\left(x>1\right)\phantom{\rule{0ex}{0ex}}x=1,\frac{1}{2}$

No roots.