Question

find the domain and range of the function. use inequality notation.
$y = -3 (0.8)^x - 3$
show your work here
hint to add inequalities (<, >, ≤, ≥), type \less\ or \greater\
domain:
range:

Explanation:

Step1: Analyze the domain of the exponential function

For the exponential function \( y = -3(0.8)^x - 3 \), the base of the exponential term is \( 0.8 \), and the exponent is \( x \). In general, for any exponential function of the form \( a^x \) (where \( a>0, a eq1 \)), the domain is all real numbers because we can raise \( a \) to any real power. So, for \( (0.8)^x \), \( x \) can be any real number. Thus, the domain of \( y = -3(0.8)^x - 3 \) is all real numbers. In inequality notation, this is \( -\infty < x < \infty \) or \( x \in (-\infty, \infty) \) (but in inequality notation as per the hint, we can write \( x \) is greater than \( -\infty \) and less than \( \infty \), but more simply, for all real numbers, the domain is \( -\infty < x < \infty \) or we can write \( x \in \mathbb{R} \) but in inequality notation, it's \( -\infty < x < \infty \)).

Step2: Analyze the range of the exponential function

First, recall the range of the basic exponential function \( a^x \) where \( 0 < a < 1 \) (like \( 0.8^x \)). The range of \( a^x \) for \( 0 < a < 1 \) is \( (0, \infty) \), meaning \( 0 < (0.8)^x < \infty \).

Now, let's transform this to find the range of \( -3(0.8)^x \). Multiply each part of the inequality \( 0 < (0.8)^x < \infty \) by \( -3 \). When we multiply an inequality by a negative number, the direction of the inequality signs flips. So:

\( 0 \times (-3) > -3(0.8)^x > \infty \times (-3) \)

Simplifying, we get \( 0 > -3(0.8)^x > -\infty \), or \( -\infty < -3(0.8)^x < 0 \).

Now, we need to find the range of \( y = -3(0.8)^x - 3 \). So we add \( -3 \) to each part of the inequality \( -\infty < -3(0.8)^x < 0 \).

Adding \( -3 \) to each part: \( -\infty - 3 < -3(0.8)^x - 3 < 0 - 3 \)

Simplifying, \( -\infty < y < -3 \). Wait, let's check that again. Wait, the original inequality for \( (0.8)^x \) is \( 0 < (0.8)^x < \infty \). Multiply by -3: \( 0 \times (-3) = 0 \), \( \infty \times (-3) = -\infty \), and the inequality signs flip, so \( -3(0.8)^x \) is between \( -\infty \) and \( 0 \), i.e., \( -\infty < -3(0.8)^x < 0 \). Then subtract 3 (or add -3) to each part: \( -\infty - 3 < -3(0.8)^x - 3 < 0 - 3 \), which is \( -\infty < y < -3 \)? Wait, no, wait. Wait, \( 0 < (0.8)^x \), so multiplying by -3 (negative) gives \( -3(0.8)^x < 0 \) (since multiplying a positive number by a negative makes it negative, and the inequality flips). And since \( (0.8)^x \) can get arbitrarily large (approaching infinity as \( x \to -\infty \)), then \( -3(0.8)^x \) can approach \( -\infty \) (since \( (0.8)^x \to \infty \) as \( x \to -\infty \), so \( -3 \times \infty = -\infty \)). Wait, maybe I messed up the direction. Let's do it step by step.

Let \( u = (0.8)^x \). Then \( u > 0 \) (since exponential functions with positive base are always positive). Then \( y = -3u - 3 \).

We need to find the range of \( y \) in terms of \( u \) where \( u > 0 \).

So, start with \( u > 0 \).

Multiply both sides by -3: \( -3u < 0 \) (because multiplying by a negative number reverses the inequality).

Now, subtract 3 from both sides: \( -3u - 3 < 0 - 3 \), so \( y < -3 \).

Now, what's the lower bound of \( y \)? As \( u \) approaches \( 0 \) from the right (since \( u > 0 \)), \( -3u \) approaches \( 0 \), so \( y = -3u - 3 \) approaches \( -3 \) from the left (since \( -3u < 0 \), so \( y = -3u - 3 < -3 \)). As \( u \) approaches \( \infty \) (since \( u = (0.8)^x \), as \( x \to -\infty \), \( u \to \infty \)), then \( -3u \) approaches \( -\infty \), so \( y = -3u - 3 \) approaches \( -\infty \). Therefore, the range of \( y \) is \( -\infty < y < -3 \), or in inequality notation, \( y < -3 \) (since it can go to \( -\infty \) and is less than -3). Wait, but let's confirm with the transformation.

Original function: \( y = -3(0.8)^x - 3 \). The base \( 0.8 \) is between 0 and 1, so \( (0.8)^x \) is a decreasing function. As \( x \to \infty \), \( (0.8)^x \to 0 \), so \( y = -3(0) - 3 = -3 \). As \( x \to -\infty \), \( (0.8)^x \to \infty \), so \( y = -3(\infty) - 3 \to -\infty \). Therefore, \( y \) can take any value less than -3, because as \( x \) increases, \( y \) approaches -3 from below (since \( (0.8)^x \) is positive, so \( -3(0.8)^x \) is negative, so \( y = -3(0.8)^x - 3 \) is more negative than -3? Wait, no: \( (0.8)^x \) is positive, so \( -3(0.8)^x \) is negative, so \( -3(0.8)^x - 3 \) is negative minus 3, so more negative. Wait, when \( x = 0 \), \( (0.8)^0 = 1 \), so \( y = -3(1) - 3 = -6 \), which is less than -3. When \( x = 1 \), \( (0.8)^1 = 0.8 \), so \( y = -3(0.8) - 3 = -2.4 - 3 = -5.4 \), still less than -3. When \( x = 10 \), \( (0.8)^{10} \approx 0.107 \), so \( y = -3(0.107) - 3 \approx -0.321 - 3 = -3.321 \), which is closer to -3 but still less than -3. As \( x \to \infty \), \( (0.8)^x \to 0 \), so \( y \to -3 \). So the range is \( y < -3 \), because \( y \) can get as close to -3 as we want (by taking \( x \) very large) but never actually reaches -3, and can go to \( -\infty \) (by taking \( x \) very small, i.e., negative and large in magnitude).

So, domain: all real numbers, so \( -\infty < x < \infty \) (or \( x \) is any real number, so in inequality notation, \( x \) is greater than \( -\infty \) and less than \( \infty \), but typically written as \( -\infty < x < \infty \) or \( x \in \mathbb{R} \), but in the required inequality notation, we can write \( x \) is greater than negative infinity and less than positive infinity, so \( -\infty < x < \infty \).

Range: \( y < -3 \), because \( y \) can be any real number less than -3 (approaching -3 from below as \( x \to \infty \) and going to \( -\infty \) as \( x \to -\infty \)).

Answer:

Domain: \( -\infty < x < \infty \)
Range: \( y < -3 \)