11.15 Topology
Exercise Find the domain of the following functions
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\(f(x) = \sqrt{1 - x^2}\)
\(\forall x \in A, \exists f(x)\) and it is unique
\(f: \mathbb{R} \rightarrow \mathbb{R}\) \(x \rightarrow f(x)\)
\(\text{Dom } f = \{ x \in \mathbb{R} : f(x) \text{ is defined} \}\)
For part (a):
\(f(x) = \sqrt{1 - x^2}\)
\(1 - x^2 \geq 0\)
\((1-x)(1+x)\geq0\Rightarrow\begin{cases}1-x\geq0\quad\text{and}\quad1+x\geq0\\ \text{or}\\ 1-x\leq0\quad\text{and}\quad1+x\leq0\end{cases}\)
\(\begin{cases}1\geq x\quad\text{and}\quad x\geq-1\Rightarrow x\in[-1,1]\\ \text{or}\\ 1\leq x\quad\text{and}\quad x\leq-1\Rightarrow\text{the intersection is empty}\end{cases}\)
\(Domf=[-1,1]\)
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\(f(x) = \frac{1}{\sqrt{1 - x^2}}\)
\(1 - x^2 > 0\)
\(\text{Dom } f = (-1, 1)\)
\(\frac{1}{0}\) is not defined in \(\mathbb{R}\)
Graphs of a function
\(g(f)=\{\left(x,f(x)\right)\in\mathbb{R}^2:x\in\text{Dom }f\}\)
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Constant function
\(f(x) = c, \quad c \in \mathbb{R}\)
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Absolute value function
\(f(x)=|x|=\begin{cases}x, & \text{if }x\geq0\\ -x, & \text{if }x<0\end{cases}\)
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\(\text{Sign function}\)
\(\text{Sign}(x)=\begin{cases}1, & x>0\\ -1, & x<0\end{cases}\)
