Find the domain and the range of the following functions.
f : x \longmapsto \log_2\left(x+3\right)
The logarithm function is defined for all positive real numbers. Therefore the domain of f equals:
\{x \in \mathbb{R} : x+3 \gt 0\}\\=\{x\in \mathbb{R}: x \gt -3\}\\=\left(-3, \infty\right)
We have:
x\in(-3,\infty)\Leftrightarrow x+3 \in (0,\infty)
and
\log_2((0,\infty))=\mathbb{R}
Thus f((-3,\infty))=\mathbb{R}.
The range of f equals (-\infty,\infty).
f :\left(-3,\infty\right) \longrightarrow \left(-\infty,\infty\right)
f : x \longmapsto \log_3\left(3x+5\right)
The logarithm function is defined for all positive real numbers. Therefore the domain of f equals:
\{x \in \mathbb{R} : 3x+5 \gt 0\}\\=\{x\in \mathbb{R}: 3x \gt -5\}\\ =\{x\in \mathbb{R}: 3x \gt -\dfrac{5}{3} \}\\=\left(-\dfrac{5}{3}, \infty\right)
We have:
x\in\left(\dfrac{-5}{3},\infty\right)\Leftrightarrow 3x+5 \in (0,\infty)
and
\log_3((0,\infty))=\mathbb{R}
Thus f\left(\left(\dfrac{-5}{3},\infty\right)\right)=\mathbb{R}.
The range of f equals (-\infty,\infty).
f :\left( -\dfrac{5}{3},\infty\right) \longrightarrow \left(-\infty,\infty\right)
f : x \longmapsto \log_5\left(x^2+1\right)
The logarithm function is defined for all positive real numbers. Therefore the domain of f equals:
\{x \in \mathbb{R} : x^2+1 \gt 0\}
Since x^2+1 is always positive, we conclude that the domain of f is \mathbb{R} =\left(-\infty, \infty\right).
We have:
x\in\mathbb{R}\Leftrightarrow x^2+1 \in [1,\infty)
and
\log_5([1,\infty))=[0,\infty)
Thus f\left(\mathbb{R}\right)=[0,\infty).
The range of f equals [0,\infty).
f :\left(-\infty,\infty\right) \longrightarrow \left[0,\infty\right)
f : x \longmapsto \log_2\left(1-x^2\right)
The logarithm function is defined for all positive real numbers. Therefore, the domain of f equals:
\{x \in \mathbb{R} : 1-x^2 \gt 0\}\\=\{x\in \mathbb{R}: x^2 \lt 1\}\\=\left(-1, 1\right)
We have:
x\in(-1{,}1)\Leftrightarrow 1-x^2 \in (0{,}1]
and
\log_2((0{,}1])=(-\infty,0]
Thus f\left((0{,}1]\right)=(-\infty,0].
The range of f equals (-\infty,0].
f :\left(-1{,}1\right)\longrightarrow\left(-\infty,0\right)
f : x \longmapsto \log_2\left(x^2-4\right)
The logarithm function is defined for all positive real numbers. Therefore, the domain of f equals:
\{x \in \mathbb{R} : x^2-4\gt 0\}\\\{x \in \mathbb{R} : x^2 \gt 4\}\\
So either:
\begin{cases} x \lt -2 \cr \text{or}\cr x \gt 2\end{cases}
Thus, the domain of f equals:
\left(-\infty,-2\right)\cup\left(2,\infty\right)
We have:
x\in(-\infty,-2)\cup(2,\infty)\Leftrightarrow x^2-4 \in (0,\infty)
and
\log_2((0,\infty))=\mathbb{R}
Thus f((-\infty,-2)\cup(2,\infty))=\mathbb{R}.
The range of f equals (-\infty,\infty).
f :\left(-\infty,-2\right)\cup\left(2,\infty\right)\longrightarrow \left(-\infty,\infty\right)
f : x \longmapsto \log_9\left(9-x^2\right)
The logarithm function is defined for all positive real numbers. Therefore, the domain of f equals:
\{x \in \mathbb{R} :9-x^2 \gt 0\}\\=\{x\in \mathbb{R}: x^2 \lt 9\}\\=\left(-3, 3\right)
We have:
x\in(-3{,}3)\Leftrightarrow 9-x^2 \in (0{,}9]
and
\log_9((0{,}9])=(-\infty,1]
Thus f((-3{,}3))=(-\infty,1].
The range of f equals (-\infty,1].
f :\left(-3{,}3\right) \longrightarrow \left(-\infty,1\right]
f : x \longmapsto \log_2\left(x^4\right)
The logarithm function is defined for all positive real numbers. Therefore, the domain of f equals:
\{x \in \mathbb{R} : x^4\gt 0\}
Since x^4 is always non-negative, only x=0 lies outside the domain. The domain is:
\mathbb{R}- \{0\} = \left(-\infty,0\right)\cup \left(0,\infty\right)
We have:
x\in(-\infty,0)\cup(0,\infty)\Leftrightarrow x^4 \in (0,\infty)
and
\log_2((0,\infty))=\mathbb{R}
Thus f((-\infty,0)\cup(0,\infty))=\mathbb{R}.
The range of f equals (-\infty,\infty).
f :\left(-\infty,0\right)\cup \left(0,\infty\right) \longrightarrow \left(-\infty,\infty\right)