Q. 5 - Calculus

Question

Write out each of the following functions as a piecewise-defined function that does not involve absolute values.

$$\begin{gathered} a) f(x)=\left|x\right| \\ b) f\left(x\right)=\left|2-3x\right| \\ c) f(x)=\left|x^2-3x-4\right| \\ d) f\left(x\right)=\left|\left|x-1\right|-1\right| \end{gathered}$$

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{Part} \left(\mathrm{a}\right): \\ \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}-\mathrm{x}& ;\mathrm{x}<0\\ \mathrm{x}& ;\mathrm{x}\ge 0\end{array}\right. \\ \mathrm{Part} \left(\mathrm{b}\right): \\ \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}3\mathrm{x}-2& ;\mathrm{x}>\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\\ 2\mathrm{x}-3& ;\mathrm{x}\le \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\end{array}\right. \\ \mathrm{Part}\left(\mathrm{c}\right): \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}-({\mathrm{x}}^2-3\mathrm{x}-4)& ;-1<\mathrm{x}<4\\ {\mathrm{x}}^2-3\mathrm{x}-4& ;\mathrm{else}\end{array}\right. \\ \mathrm{Part} \left(\mathrm{d}\right): \\ \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}-\mathrm{x}& ;\mathrm{x}<0\\ \mathrm{x}& ;0<\mathrm{x}<1\\ 2-\mathrm{x}& ;1\le \mathrm{x}<2\\ \mathrm{x}-2& ;\mathrm{x}\ge 2\end{array}\right. \end{gathered}$$

Part (a) Step 1. Given information is:

$\mathrm{f}\left(\mathrm{x}\right)=\left|\mathrm{x}\right|$

Part (a) Step 2. Piecewise Defined Function

$$\begin{gathered} \mathrm{Consider}: \\ \mathrm{f}\left(\mathrm{x}\right)=\left|\mathrm{x}\right| \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}-\mathrm{x}& ;\mathrm{x}<0\\ \mathrm{x}& ;\mathrm{x}\ge 0\end{array}\right. \end{gathered}$$

Part (b) Step 1. Given information is:

$f\left(x\right)=\left|2-3x\right|$

Part (b) Step 2. Piecewise Defined Function

$$\begin{gathered} \mathrm{Consider}: \\ \mathrm{f}\left(\mathrm{x}\right)=|2-3\mathrm{x}| \\ \mathrm{f}\left(\mathrm{x}\right)=3\mathrm{x}-2; \\ 2-3\mathrm{x}<0 \\ \mathrm{x}>\frac{2}{3} \\ \mathrm{f}\left(\mathrm{x}\right)=2-3\mathrm{x}; \\ 2-3\mathrm{x}\ge 0 \\ \mathrm{x}\le \frac{2}{3} \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}3\mathrm{x}-2& ;\mathrm{x}>\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\\ 2\mathrm{x}-3& ;\mathrm{x}\le \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\end{array}\right. \end{gathered}$$

Part (c) Step 1. Given information is:

$f\left(x\right)=\left|x^2-3x-4\right|$

Part (c) Step 2. Piecewise Defined Function

$$\begin{gathered} \mathrm{Consider}: \\ \mathrm{f}\left(\mathrm{x}\right)=|{\mathrm{x}}^2-3\mathrm{x}-4| \\ \mathrm{f}\left(\mathrm{x}\right)=-({\mathrm{x}}^2-3\mathrm{x}-4); {\mathrm{x}}^2-3\mathrm{x}-4<0 \\ \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-3\mathrm{x}-4; {\mathrm{x}}^2-3\mathrm{x}-4\ge 0; -1<\mathrm{x}<4 \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}-({\mathrm{x}}^2-3\mathrm{x}-4)& ;-1<\mathrm{x}<4\\ {\mathrm{x}}^2-3\mathrm{x}-4& ;\mathrm{else}\end{array}\right. \end{gathered}$$

Part (d) Step 1. Given information is:

$f\left(x\right)=\left|\left|x-1\right|-1\right|$

Part (d) Step 2. Piecewise Defined Function

$$\begin{gathered} \mathrm{Consider}: \\ \mathrm{f}\left(\mathrm{x}\right)=\left|\left|\mathrm{x}-1\right|-1\right| \\ \mathrm{Let} \mathrm{g}\left(\mathrm{x}\right)=\left|\mathrm{x}-1\right| \\ \mathrm{therefore}, \mathrm{g}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}1-\mathrm{x}& ;\mathrm{x}<1\\ \mathrm{x}-1& ;\mathrm{x}\ge 1\end{array}\right. \\ \mathrm{f}\left(\mathrm{x}\right)=-(1-\mathrm{x}-1)=\mathrm{x}; \mathrm{x}>0, \mathrm{x}<1 \\ \mathrm{f}\left(\mathrm{x}\right)=(1-\mathrm{x}-1)=-\mathrm{x}; \mathrm{x}<0 \\ \mathrm{f}\left(\mathrm{x}\right)=-(\mathrm{x}-1-1)=2-\mathrm{x}; \mathrm{x}<2, \mathrm{x}\ge 1 \\ \mathrm{f}\left(\mathrm{x}\right)=(\mathrm{x}-1-1)=\mathrm{x}-2; \mathrm{x}\ge 2 \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}-\mathrm{x}& ;\mathrm{x}<0\\ \mathrm{x}& ;0<\mathrm{x}<1\\ 2-\mathrm{x}& ;1\le \mathrm{x}<2\\ \mathrm{x}-2& ;\mathrm{x}\ge 2\end{array}\right. \end{gathered}$$