Domain And Range Of $7^x$ And $-(7^x)$ Explained

Alright guys, let's dive into the awesome world of functions and figure out the domain and range for a couple of cool exponential functions: $f(x) = -(7)^x$ and $g(x) = 7^x$. Understanding domain and range is super crucial in math, and for these guys, it's pretty straightforward once you get the hang of it. We'll break down exactly why they're similar in some ways and different in others.

Understanding Domain and Range

Before we get our hands dirty with these specific functions, let's quickly recap what domain and range actually mean. The domain of a function is simply all the possible input values (usually the 'x' values) that the function can accept without breaking. Think of it as the set of 'allowed' x-values. The range, on the other hand, is all the possible output values (usually the 'y' or 'f(x)' values) that the function can produce from those allowed inputs. It's the set of 'achievable' y-values.

Now, when we talk about exponential functions, which have the form $y = a^x$ where 'a' is a positive constant not equal to 1, there are some general rules. For a basic exponential function like $g(x) = 7^x$, the domain is typically all real numbers. This means you can plug in any number for 'x' – positive, negative, zero, fractions, you name it – and the function will give you a valid output. There are no restrictions like dividing by zero or taking the square root of a negative number that would stop you. We can write this domain as $(-\infty, \infty)$ or $\mathbb{R}$.

The range for these basic exponential functions is a bit more specific. Since 7 raised to any real power will always result in a positive number, the range of $g(x) = 7^x$ is all positive real numbers. It will never be zero or negative. So, the range is $(0, \infty)$. Even if you input a huge negative number for 'x', like $7^{-1000}$, the result is a very, very small positive number (close to zero), but still positive. If you input a huge positive number, the output gets incredibly large, heading towards infinity.

Analyzing $g(x) = 7^x$

Let's focus on $g(x) = 7^x$. As we just discussed, this is a classic exponential growth function.

• Domain of $g(x)$: For $g(x) = 7^x$, you can substitute any real number for 'x'. There are no mathematical operations that restrict the input. You can raise 7 to the power of 5, -2, 0.5, or any other real number. Therefore, the domain of $g(x)$ is all real numbers. In interval notation, this is represented as $(-\infty, \infty)$. This means we are completely free to choose any 'x' value we want to work with for this function. It's a wide-open playing field!

• Range of $g(x)$: Now, let's think about the outputs. When you raise 7 to any power 'x', the result is always positive. For example, $7^2 = 49$, $7^0 = 1$, and $7^{-3} = 1/343$. Notice that none of these results are zero or negative. As 'x' gets smaller and smaller (approaching negative infinity), $7^x$ gets closer and closer to zero, but it never actually reaches zero. As 'x' gets larger and larger (approaching positive infinity), $7^x$ grows without bound, approaching infinity. So, the range of $g(x)$ is all positive real numbers. In interval notation, this is $(0, \infty)$. The function can output any number greater than zero, but it can never output zero or any negative number. It's always on the positive side of the number line.

Analyzing $f(x) = -(7)^x$

Now, let's bring in our other function, $f(x) = -(7)^x$. This function looks very similar to $g(x) = 7^x$, but there's a negative sign out in front. This negative sign is a game-changer for the output, but not for the input.

• Domain of $f(x)$: Just like with $g(x)$, the part of the function that determines the domain is the exponential term $(7)^x$. Since we can input any real number for 'x' into $(7)^x$, the negative sign in front doesn't change that. We can still calculate $-(7)^2$, $-(7)^0$, or $-(7)^{-3}$. So, the domain of $f(x)$ is also all real numbers. It's the same as $g(x)$: $(-\infty, \infty)$. The domain remains unaffected by the negative multiplier.

• Range of $f(x)$: This is where things get interesting and differ from $g(x)$. Remember that the outputs of $(7)^x$ are always positive. When we multiply these positive outputs by -1 (because of the negative sign in $f(x) = -(7)^x$), the results become negative. For example, if $x=2$, $g(2) = 7^2 = 49$, but $f(2) = -(7^2) = -49$. If $x=0$, $g(0) = 7^0 = 1$, but $f(0) = -(7^0) = -1$. If $x=-3$, $g(-3) = 7^{-3} = 1/343$, but $f(-3) = -(7^{-3}) = -1/343$.

Since $(7)^x$ can produce any positive number, $-(7)^x$ can produce any negative number. As $(7)^x$ approaches 0 (from the positive side) when x goes to negative infinity, $-(7)^x$ approaches 0 (from the negative side). It gets closer and closer to zero, but never reaches it. As $(7)^x$ approaches infinity when x goes to positive infinity, $-(7)^x$ approaches negative infinity. Therefore, the range of $f(x)$ is all negative real numbers. In interval notation, this is $(-\infty, 0)$. The function can output any number less than zero, but it can never output zero or any positive number. The negative sign has essentially flipped the graph of $g(x)$ upside down.

Conclusion: Domain and Range Comparison

So, let's put it all together.

• For $g(x) = 7^x$:

  • Domain: $(-\infty, \infty)$ (all real numbers)
  • Range: $(0, \infty)$ (all positive real numbers)

• For $f(x) = -(7)^x$:

  • Domain: $(-\infty, \infty)$ (all real numbers)
  • Range: $(-\infty, 0)$ (all negative real numbers)

Comparing these, we can see that both $f(x)$ and $g(x)$ have the exact same domain: all real numbers. However, their ranges are completely different. $g(x)$ outputs only positive numbers, while $f(x)$ outputs only negative numbers.

This leads us to the correct statement: $f(x)$ and $g(x)$ have the same domain but different ranges. It's a classic example of how a simple sign change can dramatically affect the output of a function while leaving the possible inputs untouched. Pretty neat, huh guys?