Chapter 6.1, Problem 34WE

To determine

The value of $\left(x\right)$ for which the expression $\sqrt{4-x}$ represents a real number.

Answer to Problem 34WE

For $x\le 4$ the expression represents a real number.

Explanation of Solution

Given information: The given expression is,

  $\sqrt{4-x}\mathrm{......}\left(1\right)$

Calculation: From (1) for $\left(x\right)$ belongs to real, the argument $4-x\ge 0$ .

  $\begin{array}{c}4-x\ge 0\text{ gives,}\\ x\le 4\end{array}$

Hence for $x\le 4$ the expression represents a real number.

To determine

The value of $\left(x\right)$ for which the expression $\sqrt{4-x^2}$ represents a real number.

Answer to Problem 34WE

For $-2\le x\le 2$ the expression represents a real number.

Explanation of Solution

Given information: The given expression is,

  $\sqrt{4-x^2}\mathrm{......}\left(2\right)$

Calculation: From (2) for $\left(x\right)$ belongs to real, the argument $4-x^2\ge 0$ .

  $\begin{array}{c}4-x^2\ge 0\text{ gives}\\ x^2-4\le 0\\ \left(x-2\right)\left(x+2\right)\le 0\\ -2\le x\le 2\end{array}$

Hence for $-2\le x\le 2$ the expression represents a real number.

To determine

The value of $\left(x\right)$ for which the expression $\sqrt[3]{4-x^2}$ represents a real number.

Answer to Problem 34WE

For any $\left(x\right)$ belongs to real the expression represents a real number.

Explanation of Solution

Given information: The given expression is,

  $\sqrt[3]{4-x^2}\mathrm{......}\left(3\right)$

Calculation: From (3) for any $\left(x\right)$ belongs to real, the expression $\sqrt[3]{4-x^2}$ will represent a real number because the argument of function is defined for all $\left(x\right)$ belongs to real if it has odd index.

To determine

The value of $\left(x\right)$ for which the expression $\sqrt{4+x^2}$ represents a real number.

Answer to Problem 34WE

For $\left(x\right)$ belongs to real the expression represents a real number.

Explanation of Solution

Given information: The given expression is,

  $\sqrt{4+x^2}\mathrm{......}\left(4\right)$

Calculation: From (4) for $\left(x\right)$ belongs to real, the argument $4+x^2\ge 0$ .

  $\begin{array}{c}\sqrt{4+x^2}\ge 0\text{ gives}\\ x\in R\mathrm{......}\left\{\begin{array}{l}\text{As }x\text{ is real so }\\ 4+x^2\text{ will be real}\end{array}\right\}\end{array}$

Hence for $\left(x\right)$ belongs to real the expression represents a real number.