Exercise 1.1.39 (39-46)

Question

Find the domain of the function.
\begin{enumerate}
\item[39.] $f(x)=\frac{x+4}{x^2-9}$
\item[40.] $f(x)=\frac{x^2+1}{x^2+4 x-21}$
\item[41.] $f(t)=\sqrt[3]{2 t-1}$
\item[42.] $g(t)=\sqrt{3-t}-\sqrt{2+t}$
\item[43.] $h(x)=\frac{1}{\sqrt[4]{x^2-5 x}}$
\item[44.] $f(u)=\frac{u+1}{1+\frac{1}{u+1}}$
\item[45.] $F(p)=\sqrt{2-\sqrt{p}}$
\item[46.] $h(x)=\sqrt{x^2-4 x-5}$
\end{enumerate}

Khagan Gulusoy · Accepted Answer

\begin{enumerate} \item [39.] Since the denominator cannot be equal to zero, we have: %$$x^2-9 eq 0 \implies x^2 eq 9\implies x otin \{\pm 3\}.$$ \begin{align*} \forall x \in \mathbb{R}:\quad &x^2-9 eq 0\ &\iff x^2 eq 9\ &\iff x eq 3 ext{ and } x eq -3. \end{align*} %In other words, $x\in (-\infty, -3)\cup (-3,3)\cup (3, +\infty);$ In other words, the domain of $f$ is $\{x \in \mathbb{R} \mid x eq -3,3\} = (-\infty, -3)\cup (-3,3)\cup (3, +\infty)$. \item [40.] Since the denominator $x^2+4x-21$ cannot be equal to zero, we have: \begin{align*} \forall x \in \mathbb{R}:\quad &x^2+4x-21 eq 0\ &\iff (x+2)^2-5^2 eq 0\ &\iff (x+2-5)(x+2+5) eq 0\ &\iff (x-3)(x+7) eq 0\ &\iff x eq -7 ext{ and } x eq 3. \end{align*} In other words, the domain of $f$ is $\{x \in \mathbb{R} \mid x eq -7,3\} = (-\infty, -7)\cup (-7,3)\cup (3, +\infty)$. \item [41.] Since any real number can be a term under a cube root, the domain is $(-\infty,+\infty)$. \item [42.] Since the term under any even-powered root must be bigger than or equal to zero, we have: $$\forall t \in \mathbb{R}: \quad \left\{ \begin{array}{rcl} 3-t\geq 0\ 2+t\geq 0 \end{array} ight. \iff \left\{ \begin{array}{rcl} t\leq 3\ t\geq -2 \end{array} ight. %\iff t\in [-2, \ 3]. $$ In other words, the domain of $g$ is $\{t \in \mathbb{R} \mid t \geq -2 ext{ or } t \leq 3\} = [-2, \ 3]$. \item [43.] Since the term under any even-powered root must be bigger or equal to zero and the denominator must not be equal to zero, we have: \begin{align*} \forall x \in \mathbb{R}:\quad &x^2-5x> 0\ &\iff x(x-5)> 0\ &\iff x>5 ext{ or } x <0. \end{align*} In other words, the domain of $h$ is $\{x \in \mathbb{R} \mid x<0 ext{ or } x>5 \} = (-\infty, 0)\cup (5, +\infty)$. \item [44.] Since denominators cannot be equal to zero, we have: $$\forall u \in \mathbb{R}: \quad \left\{ \begin{array}{rcl} 1+\frac{1}{u+1} eq 0\ u+1 eq 0 \end{array} ight. \iff \left\{ \begin{array}{rcl} \frac{1}{u+1} eq -1\ u eq -1 \end{array} ight. \iff \left\{ \begin{array}{rcl} u+1 eq -1\ u eq -1 \end{array} ight. \iff \left\{ \begin{array}{rcl} u eq -2\ u eq -1 \end{array} ight. $$ In other words, the domain of $f$ is $\{u \in \mathbb{R} \mid u eq -2,-1\} = (-\infty, -2)\cup (-2,-1)\cup (-1, +\infty)$. \item[45.] In this example, you can see that we have 2 squared roots. That is why the domain is coming down to the system of equations: $$\forall p \in \mathbb{R}: \quad \left\{ \begin{array}{rcl} 2-\sqrt{p}\geq 0\ \sqrt{p}\geq 0 \end{array} ight. \iff \left\{ \begin{array}{rcl} \sqrt{p}\leq 2\ \sqrt{p}\geq 0 \end{array} ight. \iff \left\{ \begin{array}{rcl} p\leq 4\ p\geq 0 \end{array} ight. %\iff p\in [0,\ 4]. $$ In other words, the domain of $F$ is $\{p \in \mathbb{R} \mid 0\leq p\leq 4 \} = [0,4]$. \item [46.] The term under any even-powered root must be bigger than or equal to zero. Since $x^2-4x-5=(x-5)(x+1),$ the domain of $h$ is $\{x \in \mathbb{R}\mid x\leq -1 ext{ or } x\geq 5\} = (-\infty, -1]\cup [5, +\infty)$. \end{enumerate}

Exercise 1.1.39 (39-46)

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Exercise 1.1.39 (39-46)

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