Q. 1 - Calculus

Question

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. 

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

Step-by-Step Solution

Answer

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

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

Step 1. Given information

We have to find the roots of the following functions :

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

Step 2. Finding roots.

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

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 $$\begin{gathered} 4-x^2=0 \\ x=+2 \end{gathered}$$

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

Let 2x-1=0

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

Let $$\begin{gathered} 2x^2+x-3=0(x>1) \\ x=1,\frac{1}{2} \end{gathered}$$

No roots.