Question

the table below shows some points on the exponential function $f(x)$. find the domain and range of the function. use inequality notation.

$x$	$f(x)$
1	$\frac{7}{3}$
2	$\frac{7}{9}$

show your work here

hint: to add inequalities ($<, >, \leq, \geq$), type \less\ or \greater\

domain:

range:

Explanation:

Step1: Determine the Domain

The domain of a function is the set of all possible input values (x - values). From the table, the x - values are 0, 1, 2. But since it's an exponential function, the domain of an exponential function \(y = a\cdot b^{x}\) (where \(a\) and \(b\) are constants) is all real numbers. However, looking at the given table, the x - values start at 0 and can be any real number (since exponential functions are defined for all real x). But from the table's x - values (0, 1, 2), we can see that the domain in terms of the function's definition (exponential function) has \(x\in\mathbb{R}\), but if we consider the pattern, the function is defined for all real numbers. But let's check the general form of exponential functions. The domain of an exponential function \(f(x)=a\cdot b^{x}\) is all real numbers, so \(x\) can be any real number, so in inequality notation, the domain is \(x\in(-\infty,\infty)\) or \( -\infty < x < \infty\). But wait, the table has x = 0,1,2, but exponential functions are defined for all real x. Wait, maybe the function is \(f(x)=7\cdot(\frac{1}{3})^{x}\), let's check: when x = 0, \(7\cdot(\frac{1}{3})^{0}=7\), x = 1, \(7\cdot(\frac{1}{3})^{1}=\frac{7}{3}\), x = 2, \(7\cdot(\frac{1}{3})^{2}=\frac{7}{9}\), which matches the table. So the function is \(f(x)=7\cdot(\frac{1}{3})^{x}\), which is an exponential function with base \(\frac{1}{3}\) (between 0 and 1, so it's a decreasing exponential function). The domain of any exponential function \(y = a\cdot b^{x}\) (where \(b>0,b
eq1\)) is all real numbers. So the domain is all real numbers, so in inequality notation, \( -\infty < x < \infty\) or \(x\in(-\infty,\infty)\).

Step2: Determine the Range

The range of a function is the set of all possible output values (f(x) - values). For the exponential function \(f(x)=7\cdot(\frac{1}{3})^{x}\), since \(b = \frac{1}{3}\) (0 < b < 1), as \(x\to\infty\), \((\frac{1}{3})^{x}\to0\), so \(f(x)\to0\). As \(x\to-\infty\), \((\frac{1}{3})^{x}\to\infty\) (since \(b^{x}\) for \(0 < b < 1\) and \(x\to-\infty\) is \(\frac{1}{b^{|x|}}\to\infty\)), so \(f(x)=7\cdot(\frac{1}{3})^{x}\to\infty\). But since \((\frac{1}{3})^{x}>0\) for all real x, then \(7\cdot(\frac{1}{3})^{x}>0\). So the range is \(y>0\) or \(0 < f(x) < \infty\). Let's check the table's f(x) values: 7, \(\frac{7}{3}\), \(\frac{7}{9}\), which are all positive and decreasing towards 0 as x increases. So the range is \(0 < y < \infty\) (or \(0 < f(x) < \infty\)).

Wait, but let's re - evaluate the domain. The domain of an exponential function is all real numbers, so in inequality notation, the domain is \( -\infty < x < \infty\) (or \(x\in(-\infty,\infty)\)). The range: since the function is \(f(x)=7\cdot(\frac{1}{3})^{x}\), and \((\frac{1}{3})^{x}>0\) for all real x, then \(7\cdot(\frac{1}{3})^{x}>0\), and as x approaches \(-\infty\), \(f(x)\) approaches \(\infty\), and as x approaches \(\infty\), \(f(x)\) approaches 0. So the range is \(0 < f(x) < \infty\) (or \(0 < y < \infty\)).

Answer:

Domain: \(-\infty < x < \infty\)
Range: \(0 < f(x) < \infty\)