Chapter 6.1, Problem 32E

To determine

To sketch the region in the $xy\text{ plane}$ defined by the inequalities $x-2y^2\ge 0,\text{ 1-x-}\left|y\right|\ge 0$ and to find its area.

Answer to Problem 32E

The domain of the function $f\left(x\right)=\frac{\sqrt[3]{2x+1}}{\sqrt[3]{2x}+2}$ is $\left(0,\infty \right)$

Explanation of Solution

Given information: Consider the expression $x-2y^2\ge 0,\text{ 1-x-}\left|y\right|\ge 0$

Calculation:

  $x-2y^2\ge 0$

  $\begin{array}{l}\Rightarrow x\ge 2y^2\\ \Rightarrow x=2y^2\text{ (boundery line)}\\ Now\text{ consider }\\ \text{1-x-}\left|y\right|\ge 0\\ \Rightarrow \text{1-}\left|y\right|\ge \text{x}\\ \Rightarrow \text{x}\le \text{1-}\left|y\right|\\ To\text{ get rid of the absolute value split the above equation into two parts}\\ \text{x}\le \text{1-y},\text{ if y}\ge 0\\ \text{x}\le \text{1-(-y)},\text{ if y}<0\\ Thebounderylines\text{ (solid lines) will be these as equations x=1-y \& x=1+y}\\ \text{and the shaded area is to the left since the inequality is "x less than}\mathrm{..}\text{"}\end{array}$