Injective Function? Proving $y = \sqrt{x^2 + 1} - X$

$$

Hey guys! Today, we're diving into a fun little problem: determining whether the function $y = \sqrt{x^2 + 1} - x$ is injective, also known as one-to-one. This is a classic question in calculus and real analysis, and understanding injectivity is super important for grasping other concepts like invertibility. So, let's break it down step by step and make sure we all get it!

Understanding Injective Functions

Before we jump into the nitty-gritty, let's quickly recap what an injective function actually is. An injective function, or a one-to-one function, is a function where each element of the range corresponds to exactly one element in the domain. In simpler terms, if $f(a) = f(b)$, then $a$ must equal $b$. Think of it like a perfect matching game – each input has its own unique output, and no two inputs share the same output. This concept is crucial in many areas of mathematics, especially when dealing with inverse functions and mappings. To really grasp this, it's helpful to visualize injective functions graphically. If you draw a horizontal line anywhere on the graph of an injective function, it should intersect the graph at most once. This is known as the horizontal line test. If the line intersects more than once, it means that there are multiple $x$ values that produce the same $y$ value, which violates the one-to-one property. Understanding this graphical representation can provide an intuitive check when you're working with functions and trying to determine their injectivity. Moreover, the concept of injectivity ties directly into the existence of inverse functions. A function must be injective to have an inverse, because if multiple inputs mapped to the same output, the inverse would not be a function (it would be a relation, but not a function, because a function requires each input to have a unique output). So, when we determine that $y = \sqrt{x^2 + 1} - x$ is injective, we're also implicitly confirming that it has an inverse function, which is another important property to consider. Knowing these foundational elements makes the process of proving injectivity not just a mechanical exercise but a deeper understanding of functional relationships.

Methods to Prove Injectivity

There are a couple of main ways we can prove that a function is injective. Let's go over them briefly before applying them to our specific function.

1. The Direct Method

The direct method is exactly what the definition suggests. We assume that $f(a) = f(b)$ for some $a$ and $b$ in the domain, and then we algebraically manipulate the equation to show that $a = b$. This method is straightforward and often the first one to try, especially for simpler functions. The key here is meticulous algebraic manipulation. You need to carefully work through the equations, applying valid operations step-by-step to isolate the variables and show their equality. Any mistake in the algebraic process can lead to a wrong conclusion, so double-checking each step is critical. For instance, squaring both sides of an equation or multiplying by a variable expression requires extra caution because these operations can sometimes introduce extraneous solutions or obscure the injectivity. By directly proving that $a$ must equal $b$ whenever $f(a)$ equals $f(b)$, you are essentially demonstrating that each output of the function corresponds to only one unique input, which is the essence of injectivity. The direct method's strength lies in its clear and logical structure, providing a concrete pathway to establish the one-to-one property.

2. Using the Derivative

If our function is differentiable (which means we can find its derivative), we can use calculus to help us out. If the derivative of a function, $f'(x)$, is either strictly positive or strictly negative over its entire domain, then the function is monotonic (either strictly increasing or strictly decreasing) and therefore injective. Think about it: if a function is always going up or always going down, it can't possibly have the same y-value for two different x-values. This is a powerful technique, especially for functions that are continuous and differentiable. Calculating the derivative is often the first step, but the real insight comes from analyzing the sign of the derivative across the function's domain. If the derivative is always positive, the function is strictly increasing; if it's always negative, the function is strictly decreasing. Either of these conditions guarantees that the function is injective. However, you need to be thorough in analyzing the derivative's sign. Are there any points where the derivative is zero or undefined? These points can be critical in determining the intervals where the function is increasing or decreasing. In some cases, the function might be injective only over certain intervals and not over its entire domain. Therefore, a careful analysis of the derivative’s sign, accounting for any critical points, is essential to correctly apply this method and confidently assert the function’s injectivity. Understanding this method also reinforces the close relationship between calculus and the properties of functions, showcasing how derivatives can provide valuable insights into a function’s behavior.

Proving Injectivity of $y = \sqrt{x^2 + 1} - x$

Okay, let's get our hands dirty and prove whether our function $y = \sqrt{x^2 + 1} - x$ is injective. We'll start with the direct method because it's often the most straightforward.

1. Direct Method

Assume that $f(a) = f(b)$ for some real numbers $a$ and $b$. This means:

$$\sqrt{a^2 + 1} - a = \sqrt{b^2 + 1} - b$$

Now, we need to manipulate this equation to show that $a = b$. Let's start by isolating the square roots:

$$\sqrt{a^2 + 1} = \sqrt{b^2 + 1} + a - b$$

Next, we'll square both sides to get rid of the square roots:

$$(\sqrt{a^2 + 1})^2 = (\sqrt{b^2 + 1} + a - b)^2$$

$$a^2 + 1 = (b^2 + 1) + 2(a - b)\sqrt{b^2 + 1} + (a - b)^2$$

Expand the right side:

$$a^2 + 1 = b^2 + 1 + 2(a - b)\sqrt{b^2 + 1} + a^2 - 2ab + b^2$$

Now, let's simplify by canceling out terms and rearranging:

$$0 = 2b^2 - 2ab + 2(a - b)\sqrt{b^2 + 1}$$

Divide by 2:

$$0 = b^2 - ab + (a - b)\sqrt{b^2 + 1}$$

Now, factor out a $(b - a)$:

$$0 = -a(b - a) + (a - b)\sqrt{b^2 + 1}$$

$$0 = (a - b)(-\sqrt{b^2 + 1} - b)$$

This gives us two possibilities:

1. $a - b = 0$, which means $a = b$.
2. $-\sqrt{b^2 + 1} - b = 0$, which means $\sqrt{b^2 + 1} = -b$.

Let's analyze the second possibility. If $\sqrt{b^2 + 1} = -b$, squaring both sides gives us:

$$b^2 + 1 = b^2$$

$$1 = 0$$

This is a contradiction! So, the second possibility is not valid. Therefore, the only valid solution is $a = b$. This elegantly demonstrates that the function is injective using the direct method. By assuming $f(a) = f(b)$ and meticulously working through the algebra, we've shown that $a$ must indeed equal $b$. This confirms that each output of the function corresponds to a unique input, satisfying the very definition of injectivity. This rigorous proof leaves no room for doubt and showcases the power of the direct method when applied correctly. Furthermore, the contradiction we reached in the second possibility is a crucial point to emphasize. It highlights the importance of scrutinizing each potential solution and considering any constraints that might invalidate certain outcomes. This level of detail is what makes a mathematical argument robust and convincing. The successful application of the direct method here not only answers the specific question about injectivity but also reinforces the problem-solving skills necessary for tackling more complex mathematical challenges.

2. Using the Derivative

Now, let's confirm our result using the derivative method. First, we need to find the derivative of $f(x) = \sqrt{x^2 + 1} - x$.

$$f'(x) = \frac{d}{dx}(\sqrt{x^2 + 1} - x)$$

Using the chain rule, the derivative of $\sqrt{x^2 + 1}$ is:

$$\frac{1}{2\sqrt{x^2 + 1}} imes 2x = \frac{x}{\sqrt{x^2 + 1}}$$

And the derivative of $-x$ is simply $-1$. So:

$$f'(x) = \frac{x}{\sqrt{x^2 + 1}} - 1$$

Now, we need to determine the sign of $f'(x)$. Let's rewrite $f'(x)$ with a common denominator:

$$f'(x) = \frac{x - \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}}$$

The denominator, $\sqrt{x^2 + 1}$, is always positive. So, the sign of $f'(x)$ depends on the numerator, $x - \sqrt{x^2 + 1}$.

Notice that $\sqrt{x^2 + 1}$ is always greater than $|x|$, which means it's always greater than $x$. Therefore, $x - \sqrt{x^2 + 1}$ is always negative. This critical observation simplifies our analysis significantly. Knowing that $\sqrt{x^2 + 1}$ is inherently larger than $|x|$ allows us to immediately conclude that the numerator, and hence the entire derivative, is negative. This avoids the need for more complex algebraic manipulations or sign charts. This highlights the importance of recognizing key relationships within the expression, as it can lead to a more elegant and direct solution. Since the numerator is always negative and the denominator is always positive, the entire derivative is always negative. This establishes that the function is strictly decreasing across its entire domain. A strictly decreasing function, by definition, is injective because it never turns around and therefore never repeats a y-value for different x-values. This aligns perfectly with our finding using the direct method, reinforcing our confidence in the injectivity of the function. Furthermore, the fact that the derivative is always negative also provides additional insights into the function's behavior. It tells us not only that the function is injective but also that it's constantly decreasing, which could be useful for other analytical tasks, such as finding limits or identifying asymptotes. This demonstrates how understanding the properties of the derivative can unlock a wealth of information about the original function.

Since $f'(x)$ is always negative, the function is strictly decreasing and therefore injective. This confirms our result from the direct method!

Conclusion

So, there you have it! We've shown using two different methods – the direct method and the derivative method – that the function $y = \sqrt{x^2 + 1} - x$ is indeed injective (one-to-one). Isn't that cool? Understanding injectivity is a fundamental concept in mathematics, and mastering these proof techniques will definitely help you tackle more complex problems down the road. Keep practicing, and you'll be a pro in no time!

Remember, the key to math is understanding the underlying concepts and practicing different techniques. Whether it's the direct algebraic manipulation or the insightful use of calculus with derivatives, each method provides a unique lens through which we can understand and verify mathematical properties. So, keep exploring, keep questioning, and most importantly, keep enjoying the journey of mathematical discovery! You guys rock!