Is Function H Decreasing? Domain Intervals Explained

What's up, math lovers! Today we're diving deep into the wild world of functions, specifically looking at a piecewise function, $h(x)$, and trying to figure out where it's taking a nosedive – you know, where it's decreasing. This is super important stuff, guys, because understanding if and where a function goes down helps us predict its behavior, find minimums, and just generally get a better grip on its mathematical personality. We're talking about an interval of the domain, so let's get our thinking caps on and break down this function piece by piece.

Our function, $h(x)$, is a bit of a chameleon. It changes its tune depending on the value of $x$. For all you cool cats where $x$ is less than 1 ($x<1$), $h(x)$ is defined as $2^x$. Now, $2^x$ is a classic exponential function, and if you’ve ever graphed $y=2^x$, you know it’s always on the rise. As $x$ gets bigger, $2^x$ gets way bigger, and it never, ever goes down. So, for the first part of our domain, (-rac{\infty}{, 1), our function is strictly increasing. This immediately tells us that option A, $(-\infty, \infty)$, and option C, "The function is increasing only," might be in the running, but we can't jump to conclusions yet because we haven't analyzed the second piece.

Now, let’s switch gears and look at the second part of our function, which kicks in when $x$ is greater than or equal to 1 ($x geq 1$). Here, $h(x)$ transforms into $\sqrt{x+3}$. This is a square root function, shifted and stretched. Think about the basic square root function, $\sqrt{x}$. As $x$ increases, $\sqrt{x}$ also increases, though at a slower pace. Our function $\sqrt{x+3}$ is just $\sqrt{x}$ shifted 3 units to the left. Since the domain for this part starts at $x=1$ and goes up to positive infinity $(1, \infty)$, we need to see what happens to $\sqrt{x+3}$ as $x$ increases in this interval. If $x$ increases, $x+3$ increases, and therefore $\sqrt{x+3}$ also increases. So, for $x geq 1$, our function is also increasing. This is a crucial piece of the puzzle, guys.

So, what have we discovered? For $x<1$, $h(x)$ is increasing. For $x geq 1$, $h(x)$ is also increasing. This means that across its entire domain, $(-\infty, \infty)$, the function $h(x)$ is always increasing. It never takes a dip, it never goes down. This brings us back to our options. Option A, $(-\infty, \infty)$, is incorrect because it implies there might be a decreasing interval within the entire domain, which we've shown isn't the case. Option D, $(1, \infty)$, is also incorrect because that's an interval where the function is increasing. This leaves us with option C: "The function is increasing only." Based on our analysis of both pieces of the function, this is the correct conclusion. We've meticulously checked both parts of the domain, and in every single part, $h(x)$ is on the way up. So, to answer the question directly: Over which interval of the domain is function $h$ decreasing? The answer is, nowhere! It's always increasing.

Let's solidify this by thinking about the graphs, guys. For $x<1$, we have the graph of $y=2^x$. This is an exponential curve that starts low on the left and shoots up rapidly as it moves to the right. Crucially, it never turns downwards. As $x$ approaches 1 from the left, $2^x$ approaches $2^1=2$. So, we have this upward-sloping curve heading towards the point (1, 2). For $x geq 1$, we have the graph of $y=\sqrt{x+3}$. Let's see what happens at the boundary, $x=1$. When $x=1$, $h(1) = \sqrt{1+3} = \sqrt{4} = 2$. So, the second piece starts exactly where the first piece left off, at the point (1, 2). Now, as $x$ increases from 1, say to $x=6$, $h(6) = \sqrt{6+3} = \sqrt{9} = 3$. The function value increases from 2 to 3. If we take $x=13$, $h(13) = \sqrt{13+3} = \sqrt{16} = 4$. Again, as $x$ increases, the function value increases. The graph of $\sqrt{x+3}$ for $x geq 1$ is the upper half of a parabola opening to the right, starting at (1, 2) and curving upwards. It continuously rises. Since both pieces of the function are strictly increasing on their respective domains, and they connect seamlessly at $x=1$, the entire function $h(x)$ is strictly increasing over its entire domain of $(-\infty, \infty)$. Therefore, there is no interval where the function $h$ is decreasing. The question asks specifically for an interval where it is decreasing, and since it never decreases, the correct characterization of the function's behavior regarding decrease is that it only increases.

To be super clear, when we talk about a function being decreasing on an interval, we mean that for any two numbers $a$ and $b$ in that interval, if $a < b$, then $h(a) > h(b)$. In our case, let's pick any interval. If the interval is entirely within $x<1$, say $(0, 1)$, then for any $a, b$ such that $0 leq a < b < 1$, we have $2^a < 2^b$, so $h(a) < h(b)$. The function is increasing. If the interval is entirely within $x geq 1$, say $(1, 5)$, then for any $a, b$ such that $1 leq a < b leq 5$, we have $\sqrt{a+3} < \sqrt{b+3}$, so $h(a) < h(b)$. The function is increasing. What if the interval spans across $x=1$? Say, we pick $a=0$ and $b=2$. We know $a < b$. $h(a) = h(0) = 2^0 = 1$. $h(b) = h(2) = \sqrt{2+3} = \sqrt{5}$. Since $1 < \sqrt{5}$ (because $1^2 = 1$ and $(\sqrt{5})^2 = 5$), we have $h(a) < h(b)$. The function is still increasing. This pattern holds true for any choice of $a < b$ across the entire domain. Thus, there is no interval where $h(a) > h(b)$ when $a < b$. The function is exclusively increasing. This confirms that option C, "The function is increasing only," is the correct response to the question about where it is decreasing because it implies there is no such decreasing interval. It’s all about understanding the individual components and how they behave together across the whole picture.

So, the breakdown is this: the first part, $2^x$ for $x<1$, is a classic example of an increasing function. Its derivative, $2^x \ln(2)$, is always positive. The second part, $\sqrt{x+3}$ for $x geq 1$, is also an increasing function. Its derivative is $\frac{1}{2\sqrt{x+3}}$, which is also always positive for $x > -3$, and certainly positive for $x geq 1$. Since both pieces are increasing on their respective domains, and they connect at the point $(1, 2)$ to form a continuous function, the entire function $h(x)$ is increasing over its entire domain $(-\infty, \infty)$. Therefore, there is no interval over which $h(x)$ is decreasing. The only accurate description among the choices that addresses the question of decreasing intervals is to state that the function is increasing only, implying no decreasing intervals exist. This is why option C is the definitive answer, guys. Keep practicing, and you'll master these piecewise functions in no time!