Chapter 6, Problem 40SGR

To determine

To Graph :

Graph the given inequality.

Explanation of Solution

Given:

The inequality is:

  $y>-\sqrt{x-1}+2$

Concept Used:

• Definition of nth root:
• Words : For any real numbers a and b , and any positive integer n , if $a^n=b$ then a is the nth root of b . That is $a=\sqrt[n]{b}$
• $\begin{array}{l}If\text{ n is even: }\\ \sqrt[n]{b}\ge 0\text{ and }b\ge 0\end{array}$
• $\begin{array}{l}\sqrt[n]{b}\\ Here\\ b=radicand\\ n=index\\ \sqrt{}=radicalsign\end{array}$
• For the inequality , if given y- value is greater ,then the shaded region will be above the boundary of the inequality.
• If given y -value is less , then the shaded region will be below the boundary of the inequality.

Graph:

  $\begin{array}{l}Given\\ y>-\sqrt{x-1}+2\\ \text{Since the index is 2, which is even; so the radicand is non-negative}\text{.}\\ x-1\ge 0\\ x\ge \mathrm{1.........}[\text{add 1 each side]}\\ So,\text{ for the inequality , }\\ \text{Domain= {x | x}\ge 1\text{}}\\ \text{First we draw the boundary }\\ y=-\sqrt{x-1}+2\\ \text{The graph of }y=-\sqrt{x-1}+2\text{ is the graph of }y=-\sqrt{x}\\ \text{shifted 1 unit to the right and 2 units upward , since the }\\ \text{vertex is (1,2)}\text{.}\\ \text{ince the }inequality\text{ is '}y>\text{'},the\text{ shaded region or the solution region will}\\ \text{be above the boundary line and excluding the boundary line as}\\ \text{the inequaltiy is strict}\text{.}\end{array}$