Question 3.3 • Calculus (4th Edition) • Textbookish

\begin{align*} \operatorname{Dom}(f) &= \{x:1-x^2\ge 0\} \\ &= \{x:x^2\le 1\} \\ &= [-1,1]. \end{align*}

ii)

\begin{align*} \operatorname{Dom}(f) &= \{x:1-x^2\ge 0 \land 1-\sqrt{1-x^2}\ge 0\} \\ &= [-1,1]\cap \{x:1-x^2\le 1\} \\ &= [-1,1]\cap \{x:-x^2\le 0\} \\ &= [-1,1]\cap \mathbb{R} \\ &= [-1,1]. \end{align*}

iii)

\operatorname{Dom}(f)=\{x:x-1\ne 0 \land x-2\ne 0\}=\{x:x\ne 1 \land x\ne 2\}.

iv)

\begin{align*} \operatorname{Dom}(f) &= \{x:1-x^2\ge 0 \land x^2-1\ge 0\} \\ &= [-1,1]\cap \{x:x^2\ge 1\} \\ &= [-1,1]\cap ((-\infty,-1]\cup[1,\infty)) \\ &= \{-1,1\}. \end{align*}

\begin{align*} \operatorname{Dom}(f) &= \{x:1-x\ge 0 \land x-2\ge 0\} \\ &= \{x:x\le 1 \land x\ge 2\} \\ &= \varnothing. \end{align*}

Hence, $f(x)=\varnothing$ .

Solutions