Analyzing Function Variations And Solving Equations

Hey guys, let's dive into a fun little math problem! We're going to explore the function $f(x) = x\sqrt{x} + x - 1$ and tackle two main challenges: understanding its behavior and solving a related equation. This will involve some calculus, some algebraic manipulation, and a bit of logical reasoning. Don't worry, it's not as scary as it sounds! We'll break it down step by step.

Studying the Variations of the Function $f$ on $R^+$

Our first goal is to figure out how the function $f(x) = x\sqrt{x} + x - 1$ behaves on the set of non-negative real numbers, denoted as $R^+$. This means we want to know where the function is increasing, decreasing, and any potential turning points. To do this, we'll use the power of calculus, specifically the derivative. The derivative of a function tells us about its rate of change.

First things first, let's find the derivative of $f(x)$. Remember that $\sqrt{x}$ is the same as $x^{1/2}$. So, $f(x)$ can be rewritten as $x^{3/2} + x - 1$. Now, applying the power rule of differentiation (which says that the derivative of $x^n$ is $nx^{n-1}$), we get:

$f'(x) = \frac{3}{2}x^{1/2} + 1$

Notice that the derivative, $f'(x)$, is $\frac{3}{2}\sqrt{x} + 1$. Since $x$ is in $R^+$, the square root of $x$ is always non-negative (i.e., greater than or equal to zero). Consequently, $\frac{3}{2}\sqrt{x}$ is also non-negative. Adding 1 to that makes the entire expression, $\frac{3}{2}\sqrt{x} + 1$, always positive. This is crucial because it tells us something very important about the function's behavior.

Specifically, a positive derivative means that the function is increasing. So, because $f'(x) > 0$ for all $x$ in $R^+$, we can conclude that the function $f(x) = x\sqrt{x} + x - 1$ is strictly increasing on $R^+$. This means as $x$ gets larger, $f(x)$ also gets larger, and there are no points where the function turns around and starts decreasing. This is a fundamental concept in calculus, and it's a pretty powerful tool for understanding the behavior of functions. To recap, we found the derivative, analyzed its sign, and then used that information to determine that the function is always increasing on the specified domain.

To provide a more complete analysis, let's consider the limits of f(x) as x approaches the boundaries of $R^+$. As x approaches 0, $f(x)$ approaches -1. As x approaches infinity, $f(x)$ also approaches infinity. These observations confirm that $f(x)$ covers all values from -1 (exclusive) upwards. This further supports our understanding of the function's behavior.

In short, the first part of the exercise is all about using derivatives to understand how a function changes. It allows us to characterize the nature of functions, which is a useful step in a more general study.

Showing That the Equation $x\sqrt{x} = 1 - x$ Admits a Unique Solution

Alright, now for the second part of the problem! We want to prove that the equation $x\sqrt{x} = 1 - x$ has only one solution. This might seem unrelated to our analysis of $f(x)$ at first, but it's actually very connected. Notice that the left-hand side of the equation, $x\sqrt{x}$, is closely related to our function $f(x)$.

Let's rearrange the equation a bit to see the connection more clearly. We can add $x$ to both sides, which gives us:

$x\sqrt{x} + x = 1$

Now, let's subtract 1 from both sides:

$x\sqrt{x} + x - 1 = 0$

Do you recognize the left-hand side? It's exactly our function $f(x)!$ So, we've essentially transformed the equation $x\sqrt{x} = 1 - x$ into $f(x) = 0$. Finding the solution(s) to the original equation is now equivalent to finding the root(s) of $f(x)$.

Since we know that $f(x)$ is strictly increasing on $R^+$, it can cross the x-axis (i.e., have a root) at most once. Why? Because an increasing function can only go from negative values to positive values once. Also note that we already showed that $f(x)$ is continuous. The fact that the function is increasing is a critical piece of information. If the function were not strictly increasing, it could potentially have multiple roots, but due to the strictly increasing nature, we know that there can only be one. More formally, this is an application of the Intermediate Value Theorem combined with the strictly increasing property of the function.

To prove the existence of the solution, we must show there's a value of $x$ for which $f(x) = 0$. Now, let's think about the behavior of $f(x)$ again. We know $f(x)$ is strictly increasing and continuous on $R^+$. Also, we know that when $x = 0$, $f(0) = -1$. And, we know that as x approaches infinity, $f(x)$ also approaches infinity. Since $f(x)$ is continuous, by the Intermediate Value Theorem, there must be a value of $x$ where $f(x) = 0$. This value of x is the solution to our equation.

Therefore, we have both the existence and the uniqueness of a solution. The original equation $x\sqrt{x} = 1 - x$ has a unique solution within the domain $R^+$. We have successfully shown the existence of a solution and proven its uniqueness. This illustrates a powerful application of calculus in analyzing equations, where the properties of functions such as monotonicity (being strictly increasing or decreasing) help determine the number of solutions. This is a prime example of how a deep understanding of a function's behavior can reveal crucial information about equations that are connected to it.

To summarize, we've shown that the given equation has one solution by cleverly manipulating it, relating it to our function $f(x)$, and using our knowledge of $f(x)$'s increasing behavior.

To complete this problem fully, you might want to attempt to find the actual value of the solution, although the problem only required to show that there is one. This can be done using numerical methods like the Newton-Raphson method.

Conclusion

In conclusion, we've successfully analyzed the function $f(x) = x\sqrt{x} + x - 1$ and solved a related equation. We used the derivative to understand the function's increasing nature and then applied this knowledge to prove the existence and uniqueness of a solution to the equation $x\sqrt{x} = 1 - x$. This exercise showcases how concepts from calculus, such as derivatives and the Intermediate Value Theorem, can be used together to solve problems. I hope this helps you understand the problems better! Feel free to ask any questions you might have. Keep practicing and having fun with math, everyone!