Finding The Range Of F(x) = (x-3)/(x+4)

Hey guys! Today, we're diving deep into the world of functions and tackling a question that might seem a little tricky at first glance: What is the range of $f(x) = \frac{x-3}{x+4}$? Now, before we get lost in the mathematical weeds, let's break down what we mean by the 'range' of a function. In simple terms, the range is the set of all possible output values that a function can produce. Think of it as the collection of 'y' values your function can hit. We're going to explore this together, unpack the options, and figure out the correct answer, making sure you guys feel totally confident about it. We'll be looking at the given function, $f(x) = \frac{x-3}{x+4}$, and using our math skills to pinpoint its range. So, grab your calculators, maybe a comfy seat, and let's get started on this mathematical adventure!

Understanding Functions and Their Ranges

Alright, let's kick things off by getting super clear on what we're dealing with. The function we've got here is $f(x) = \frac{x-3}{x+4}$. This is what we call a rational function, which basically means it's a fraction where both the numerator and the denominator are polynomials. In this case, the numerator is $x-3$ and the denominator is $x+4$. Now, the domain of a function is all the possible input values (the 'x' values) that you can plug into the function without breaking it. For rational functions, the main thing to watch out for is division by zero. If the denominator becomes zero, the function is undefined at that 'x' value. For our function, $f(x) = \frac{x-3}{x+4}$, the denominator $x+4$ is zero when $x = -4$. So, the domain of this function is all real numbers except for $x = -4$. We can write this as $\{x \in \mathbb{R} \mid x \neq -4\}$.

But today, our main quest is to find the range. The range, as we mentioned, is the set of all possible output values (the 'y' or $f(x)$ values). To find the range of a rational function like this, we often try to see what values the function cannot take. One common way to do this is to try and express 'x' in terms of 'y'. This means we set $y = f(x)$ and then solve for 'x'. Let's give that a shot:

$$y = \frac{x-3}{x+4}$$

Our goal here is to isolate 'x'. First, multiply both sides by $(x+4)$ to get rid of the fraction:

$$y(x+4) = x-3$$

Now, distribute the 'y' on the left side:

$$xy + 4y = x-3$$

We want to get all the terms with 'x' on one side and everything else on the other. Let's move the 'x' term from the right to the left and the $4y$ term from the left to the right:

$$xy - x = -3 - 4y$$

Now, factor out 'x' from the terms on the left:

$$x(y - 1) = -3 - 4y$$

Finally, to solve for 'x', divide both sides by $(y-1)$:

$$x = \frac{-3 - 4y}{y - 1}$$

Now, what does this expression for 'x' tell us about the possible values of 'y'? Just like with the original function where we couldn't have a denominator of zero, here, we can't have the denominator $(y-1)$ equal to zero. If $y-1 = 0$, then $y=1$. This means that $y$ can never be equal to 1. If $y$ were 1, we'd be trying to divide by zero, which is a no-go in mathematics. So, based on this manipulation, we can see that the function $f(x)$ can take on any real value except for $y=1$. This is a crucial piece of information that helps us determine the range.

Analyzing the Options Provided

We've done the heavy lifting and figured out that $y$ cannot be equal to 1. Now, let's look at the options provided to see which one matches our findings:

A. $y>1$ B. $y \neq 0$ C. $y \neq 4$ D. $y \neq 1$

Option A, $y > 1$, is too restrictive. It claims that 'y' must be greater than 1, but our analysis showed that 'y' can be any real number except 1. This means it could be less than 1 (e.g., 0, -5, etc.). So, A is out.

Option B, $y \neq 0$, suggests that 'y' can be any value except 0. Let's quickly check if $y=0$ is possible. If $f(x) = 0$, then $\frac{x-3}{x+4} = 0$. For a fraction to be zero, the numerator must be zero (as long as the denominator isn't also zero). So, $x-3=0$, which means $x=3$. Since $x=3$ is a valid input (it's not -4), $f(3) = \frac{3-3}{3+4} = \frac{0}{7} = 0$. This means $y=0$ is a possible output. Therefore, the range is not $y \neq 0$. So, B is incorrect.

Option C, $y \neq 4$, suggests that 'y' can be any value except 4. We haven't seen anything in our derivation that directly prohibits $y=4$. Let's see if $y=4$ is achievable. If $f(x)=4$, then $\frac{x-3}{x+4} = 4$. Multiplying both sides by $(x+4)$, we get $x-3 = 4(x+4)$, so $x-3 = 4x+16$. Rearranging gives $-3-16 = 4x-x$, which is $-19 = 3x$, so $x = -19/3$. Since $x = -19/3$ is a valid input (it's not -4), $y=4$ is indeed a possible output. Thus, the range is not $y \neq 4$. So, C is incorrect.

Option D, $y \neq 1$, states that 'y' can be any real number except 1. This perfectly matches our derivation when we solved for 'x' in terms of 'y' and found that the denominator $(y-1)$ could not be zero, implying $y \neq 1$. This is the correct range for the function $f(x) = \frac{x-3}{x+4}$.

Alternative Approach: Horizontal Asymptotes

For those of you who are familiar with the concepts of asymptotes, there's another slick way to think about the range of rational functions. A horizontal asymptote tells us about the behavior of the function as 'x' approaches positive or negative infinity. For a rational function of the form $f(x) = \frac{ax^n + ...}{bx^m + ...}$, where 'n' is the degree of the numerator and 'm' is the degree of the denominator:

1. If $n < m$, the horizontal asymptote is $y = 0$.
2. If $n > m$, there is no horizontal asymptote (but there might be a slant asymptote).
3. If $n = m$, the horizontal asymptote is $y = \frac{a}{b}$ (the ratio of the leading coefficients).

In our function, $f(x) = \frac{x-3}{x+4}$, the degree of the numerator is 1 (from the 'x' term) and the degree of the denominator is also 1 (from the 'x' term). Since the degrees are equal ($n=m=1$), the horizontal asymptote is the ratio of the leading coefficients. The leading coefficient in the numerator is 1 (from $1x$) and the leading coefficient in the denominator is also 1 (from $1x$). So, the horizontal asymptote is $y = \frac{1}{1} = 1$.

What does a horizontal asymptote signify for the range? It indicates a value that the function approaches but may or may not actually reach. For many rational functions, the horizontal asymptote represents a value that the function never actually outputs. In this specific case, the horizontal asymptote is $y=1$. This means that as 'x' gets very, very large (either positive or negative), the value of $f(x)$ gets closer and closer to 1, but it never quite reaches 1. This aligns perfectly with our previous algebraic method, which showed that $y$ cannot be equal to 1.

It's important to note, however, that just because $y=c$ is a horizontal asymptote, it doesn't automatically mean the function never reaches $c$. There are exceptions, especially with more complex functions or when the function's graph crosses its horizontal asymptote for finite values of 'x'. But for simple rational functions like this one, the horizontal asymptote is generally a value excluded from the range. Our algebraic method is the most rigorous way to confirm this exclusion.

So, the horizontal asymptote at $y=1$ strongly suggests that $y=1$ is not in the range of the function. This reinforces our conclusion from the algebraic manipulation.

Visualizing the Graph

Let's think about what the graph of $f(x) = \frac{x-3}{x+4}$ looks like. We know it has a vertical asymptote at $x=-4$ (where the denominator is zero) and a horizontal asymptote at $y=1$. The function will approach these asymptotes but never touch them.

For $x > -4$, as $x$ increases, the denominator $x+4$ becomes a larger positive number, and the numerator $x-3$ also increases. The ratio $\frac{x-3}{x+4}$ will approach 1 from below. For example, if $x=0$, $f(0) = -3/4$. If $x=3$, $f(3)=0$. If $x=10$, $f(10) = 7/14 = 1/2$. As $x$ gets very large, say $x=1000$, $f(1000) = 997/1004$, which is very close to 1.

For $x < -4$, as $x$ becomes more negative (e.g., $x=-5$), $f(-5) = \frac{-5-3}{-5+4} = \frac{-8}{-1} = 8$. If $x=-10$, $f(-10) = \frac{-10-3}{-10+4} = \frac{-13}{-6} = \frac{13}{6} \\approx 2.17$. As $x$ approaches $-4$ from the left, the denominator $x+4$ becomes a very small negative number, making the function's value very large and positive. As $x$ approaches $-\infty$, the function approaches 1 from above. For example, if $x=-1000$, $f(-1000) = \frac{-1003}{-996} \\approx 1.007$.

Sketching this out, you'd see two branches of a hyperbola, one in the upper-right region relative to the asymptotes and one in the lower-left. The key takeaway from the graph is that the 'y' values cover everything except the horizontal line $y=1$. The graph never intersects the line $y=1$. This visual confirmation is super helpful for understanding the range.

Conclusion: The Correct Range

After exploring the problem using algebraic manipulation and considering the concept of horizontal asymptotes, we've consistently arrived at the same conclusion: the function $f(x) = \frac{x-3}{x+4}$ can produce any real output value except for $y=1$. This is because if we set $y=1$, our attempts to solve for a corresponding 'x' value lead to an undefined expression (division by zero). Therefore, the value $y=1$ is excluded from the range.

Looking back at our options:

A. $y>1$ B. $y \neq 0$ C. $y \neq 4$ D. $y \neq 1$

Option D, $y \neq 1$, is the only one that accurately describes the set of all possible output values for the function $f(x) = \frac{x-3}{x+4}$. So, guys, the answer is D! Keep practicing these types of problems, and you'll become absolute pros at finding function ranges in no time. Math on!