Find The Function: Domain $(-\infty, \infty)$, Range $(-\infty, 4]$

Hey math whizzes and curious minds of Plastik Magazine! Today, we're diving deep into the fascinating world of functions. We've got a classic puzzle for you: which function boasts a domain stretching from negative infinity to positive infinity, and a range that caps out at positive four? This isn't just about memorizing rules, guys; it's about understanding the DNA of functions and how their graphs behave. Let's break down what domain and range really mean and then apply that knowledge to solve this head-scratcher.

Understanding the Building Blocks: Domain and Range

First off, let's get crystal clear on what domain and range are. Think of the domain as the entire set of possible input values for a function. It's all the 'x' values you can plug into the function without breaking it (like dividing by zero or taking the square root of a negative number, if we were dealing with those kinds of functions). When we see the symbol $(-\infty, \infty)$, it means the domain is all real numbers. No matter what real number you choose for 'x', the function will happily give you an output. Functions like linear functions ($f(x) = mx + b$) and quadratic functions ($f(x) = ax^2 + bx + c$, where $a \neq 0$) typically have a domain of all real numbers.

Now, let's talk about the range. The range is the set of all possible output values – the 'y' or $f(x)$ values that the function can produce. The range $(-\infty, 4]$ tells us something really specific. It means the function can output any real number from negative infinity up to and including 4. The square bracket ] at the 4 signifies that 4 itself is a possible output. This immediately tells us that the function's graph doesn't go infinitely upwards; it has a ceiling, and that ceiling is at $y=4$.

Analyzing Our Contenders: A Closer Look

We've been given four potential functions, and it's our job to put them under the microscope. Let's examine each one to see if it matches our specific domain and range requirements. Remember, we're looking for a function where any real number can be plugged in for 'x' (domain: $(-\infty, \infty)$), and the resulting 'y' values are always less than or equal to 4 (range: $(-\infty, 4]$).

1. $f(x) = x + 4$ Let's start with $f(x) = x + 4$. This is a simple linear function. What's its domain? Can we plug in any real number for 'x'? Absolutely! There are no restrictions here. So, the domain is indeed $(-\infty, \infty)$. Now, what about the range? As 'x' goes from $-\infty$ to $\infty$, what happens to $x+4$? It also goes from $-\infty$ to $\infty$. Think about it: if $x$ is a huge positive number, $x+4$ is a huge positive number. If $x$ is a huge negative number, $x+4$ is a huge negative number. Therefore, the range of $f(x) = x + 4$ is $(-\infty, \infty)$. This doesn't match our target range of $(-\infty, 4]$. So, this one's out, guys.

2. $f(x) = -4x$ Next up is $f(x) = -4x$. This is another linear function, but this time it has a negative slope. What's the domain? Just like the previous function, we can plug in any real number for 'x'. No issues there. The domain is $(-\infty, \infty)$. Now for the range. As 'x' increases, $-4x$ decreases. If $x$ is a large positive number, $-4x$ is a large negative number. If $x$ is a large negative number, $-4x$ is a large positive number. This function also covers all real numbers for its output. So, the range is $(-\infty, \infty)$. Again, this doesn't match our desired range of $(-\infty, 4]$. Moving on!

3. $f(x) = -x^2 + 4$ Alright, this one looks different: $f(x) = -x^2 + 4$. This is a quadratic function. Let's talk domain first. Can we square any real number 'x' and then subtract it from 4? Yes, we can! There are no limitations on the 'x' values we can input. So, the domain is $(-\infty, \infty)$. Perfect! Now, let's investigate the range. The key part here is the $-x^2$ term. Remember that $x^2$ is always greater than or equal to zero ($x^2 \ge 0$) for any real number 'x'. When we multiply it by -1, we get $-x^2$, which is always less than or equal to zero ($-x^2 \le 0$). So, the term $-x^2$ will always produce a non-positive value. Now, we add 4 to it: $-x^2 + 4$. Since $-x^2$ is always less than or equal to 0, the largest possible value for $-x^2 + 4$ occurs when $-x^2$ is at its maximum, which is 0. When $-x^2 = 0$ (which happens when $x=0$), the function's value is $f(0) = -(0)^2 + 4 = 4$. For any other value of 'x', $x^2$ will be positive, $-x^2$ will be negative, and $-x^2 + 4$ will be less than 4. For example, if $x=1$, $f(1) = -(1)^2 + 4 = -1 + 4 = 3$. If $x=-2$, $f(-2) = -(-2)^2 + 4 = -4 + 4 = 0$. As 'x' gets really large in either the positive or negative direction, $x^2$ gets really large, $-x^2$ gets really large negatively, and thus $-x^2 + 4$ heads towards negative infinity. So, the output values are always less than or equal to 4. This means the range is indeed $(-\infty, 4]$. Bingo! This function fits both our domain and range requirements.

4. $f(x) = 2^x + 4$ Last but not least, we have $f(x) = 2^x + 4$. This is an exponential function. Let's tackle the domain first. Can we raise 2 to the power of any real number 'x'? Yes, we can! So, the domain is $(-\infty, \infty)$. Great. Now, what about the range? The base function $2^x$ has a range of $(0, \infty)$. This means $2^x$ is always positive, but it never reaches zero, and it can grow infinitely large. When we add 4 to it, $2^x + 4$, we are shifting the entire graph upwards by 4 units. So, the outputs will always be greater than 4 ($2^x + 4 > 4$). For instance, if $x=0$, $f(0) = 2^0 + 4 = 1 + 4 = 5$. If $x=1$, $f(1) = 2^1 + 4 = 2 + 4 = 6$. As 'x' approaches negative infinity, $2^x$ approaches 0, so $2^x + 4$ approaches 4. As 'x' approaches positive infinity, $2^x$ approaches infinity, so $2^x + 4$ approaches infinity. Therefore, the range of $f(x) = 2^x + 4$ is $(4, \infty)$. This does not match our target range of $(-\infty, 4]$.

The Verdict: Case Closed!

After meticulously examining each function, we found our winner! The function that has a domain of $(-\infty, \infty)$ and a range of $(-\infty, 4]$ is $f(x) = -x^2 + 4$. This is because the $-x^2$ term ensures the outputs are always less than or equal to zero, and adding 4 shifts the maximum output to exactly 4, while the domain remains unrestricted. Keep practicing these concepts, guys, and you'll become math masters in no time! What other function mysteries can we solve next?