Injective Function? Solving Y = √(x² + 1) - X

by ADMIN 46 views
Iklan Headers

Hey guys! Let's dive into an interesting question today: Is the function y=x2+1−xy = \sqrt{x^2 + 1} - x an injective (one-to-one) function? This is a classic problem in calculus and real analysis, and understanding injectivity is crucial for many advanced mathematical concepts. So, let's break it down step-by-step and make sure we grasp every detail.

Understanding Injective Functions

First off, what exactly does it mean for a function to be injective, also known as one-to-one? Simply put, a function f is injective if it never maps two different elements in its domain to the same element in its codomain. Mathematically, this can be expressed as:

If f(a) = f(b), then a = b for all a and b in the domain of f.

In simpler terms, if two different inputs always produce different outputs, then the function is injective. Think of it like this: each input has a unique output, and no two inputs share the same output. This property is essential for the existence of an inverse function, which is another important concept in mathematics.

To really solidify this, let's consider some examples. The function f(x) = x is injective because every distinct input x produces a distinct output f(x). On the other hand, the function f(x) = x² is not injective over the real numbers because f(-2) = f(2) = 4. Two different inputs (-2 and 2) produce the same output (4). However, if we restrict the domain to non-negative real numbers, f(x) = x² becomes injective because for any non-negative a and b, if a² = b², then a = b. Understanding these fundamental concepts is key to tackling more complex problems like the one we're addressing today.

Now, with a solid grasp of what injectivity means, we can start thinking about how to approach proving whether our function, y=x2+1−xy = \sqrt{x^2 + 1} - x, is injective. We'll explore different methods, from algebraic manipulations to calculus-based approaches, to provide a comprehensive understanding of the problem.

Methods to Prove Injectivity

There are several ways to prove whether a function is injective. We'll focus on two main methods:

  1. Algebraic Method: Directly using the definition of injectivity.
  2. Calculus Method: Using the derivative of the function.

1. Algebraic Method: Directly Using the Definition

This method involves directly applying the definition of injectivity. We assume that f(a) = f(b) for some a and b in the domain and then try to show that a = b. This is often the most straightforward approach, but it can sometimes involve tricky algebraic manipulations.

So, let's start by assuming that:

a2+1−a=b2+1−b\sqrt{a^2 + 1} - a = \sqrt{b^2 + 1} - b

Our goal is to manipulate this equation to show that a = b. This might seem daunting at first, but with a few clever steps, we can simplify it. The key here is to isolate the square roots and then eliminate them by squaring. This is a common technique when dealing with equations involving square roots.

First, let's rearrange the equation to group the square root terms on one side and the linear terms on the other:

a2+1−b2+1=a−b\sqrt{a^2 + 1} - \sqrt{b^2 + 1} = a - b

Now, this looks a bit more manageable. To get rid of the square roots, we can use a classic trick: multiply both sides by the conjugate of the left-hand side. The conjugate is formed by simply changing the sign between the terms. So, the conjugate of a2+1−b2+1\sqrt{a^2 + 1} - \sqrt{b^2 + 1} is a2+1+b2+1\sqrt{a^2 + 1} + \sqrt{b^2 + 1}. Multiplying by the conjugate will allow us to use the difference of squares formula, which will eliminate the square roots.

Multiplying both sides by the conjugate, we get:

(a2+1−b2+1)(a2+1+b2+1)=(a−b)(a2+1+b2+1)(\sqrt{a^2 + 1} - \sqrt{b^2 + 1})(\sqrt{a^2 + 1} + \sqrt{b^2 + 1}) = (a - b)(\sqrt{a^2 + 1} + \sqrt{b^2 + 1})

Now, applying the difference of squares formula, (A - B)(A + B) = A² - B², we have:

(a2+1)−(b2+1)=(a−b)(a2+1+b2+1)(a^2 + 1) - (b^2 + 1) = (a - b)(\sqrt{a^2 + 1} + \sqrt{b^2 + 1})

Simplifying the left side, we get:

a2−b2=(a−b)(a2+1+b2+1)a^2 - b^2 = (a - b)(\sqrt{a^2 + 1} + \sqrt{b^2 + 1})

Now, we can factor the left side as a difference of squares:

(a−b)(a+b)=(a−b)(a2+1+b2+1)(a - b)(a + b) = (a - b)(\sqrt{a^2 + 1} + \sqrt{b^2 + 1})

At this point, we have two possibilities: either a - b = 0 or we can divide both sides by (a - b). If a - b = 0, then a = b, which is what we want to prove. So, let's consider the case where a - b ≠ 0. In this case, we can divide both sides by (a - b), giving us:

a+b=a2+1+b2+1a + b = \sqrt{a^2 + 1} + \sqrt{b^2 + 1}

Now, we have two equations:

  1. a2+1−b2+1=a−b\sqrt{a^2 + 1} - \sqrt{b^2 + 1} = a - b
  2.  a+b=a2+1+b2+1\ a + b = \sqrt{a^2 + 1} + \sqrt{b^2 + 1}

Adding these two equations together, we get:

2a=2a2+12a = 2\sqrt{a^2 + 1}

Dividing both sides by 2, we have:

a=a2+1a = \sqrt{a^2 + 1}

Squaring both sides, we get:

a2=a2+1a^2 = a^2 + 1

This simplifies to:

0=10 = 1

Which is a contradiction! This means our assumption that a - b ≠ 0 must be false. Therefore, the only possibility is that a - b = 0, which implies a = b. Thus, we have successfully shown that if f(a) = f(b), then a = b, using the algebraic method. This provides strong evidence that the function is injective.

2. Calculus Method: Using the Derivative

Another powerful method to determine injectivity involves using calculus, specifically the derivative of the function. If a function is strictly increasing or strictly decreasing over its entire domain, then it is injective. This is because a strictly increasing function always moves upwards, and a strictly decreasing function always moves downwards, so they can never have the same y-value for two different x-values.

To apply this method, we need to find the derivative of our function, y=x2+1−xy = \sqrt{x^2 + 1} - x, and analyze its sign. If the derivative is always positive or always negative, then the function is strictly monotonic (either increasing or decreasing) and therefore injective.

Let's find the derivative, dy/dx, using the chain rule and the power rule:

dydx=ddx(x2+1−x)\frac{dy}{dx} = \frac{d}{dx}(\sqrt{x^2 + 1} - x)

dydx=ddx((x2+1)1/2)−ddx(x)\frac{dy}{dx} = \frac{d}{dx}((x^2 + 1)^{1/2}) - \frac{d}{dx}(x)

Using the chain rule for the first term:

\frac{dy}{dx} = \frac{1}{2}(x^2 + 1)^{-1/2} ullet 2x - 1

Simplifying, we get:

dydx=xx2+1−1\frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 1}} - 1

Now, we need to determine the sign of this derivative. To do this, let's analyze the expression. We can rewrite the derivative as:

dydx=x−x2+1x2+1\frac{dy}{dx} = \frac{x - \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}}

The denominator, x2+1\sqrt{x^2 + 1}, is always positive. So, the sign of the derivative depends on the numerator, x - \sqrt{x^2 + 1}. We need to determine if this numerator is always positive, always negative, or sometimes positive and sometimes negative.

Notice that x2+1\sqrt{x^2 + 1} is always greater than ∣x∣|x| because we are adding 1 inside the square root. This means that x2+1>x\sqrt{x^2 + 1} > x for all x. Therefore, x - \sqrt{x^2 + 1} is always negative.

Since the numerator is always negative and the denominator is always positive, the derivative dy/dx is always negative. This means that the function y=x2+1−xy = \sqrt{x^2 + 1} - x is strictly decreasing over its entire domain (which is all real numbers).

Because the function is strictly decreasing, it is injective. This calculus-based method provides another strong confirmation that our function is indeed one-to-one.

Conclusion

So, guys, after our thorough exploration using both the algebraic and calculus methods, we can confidently conclude that the function y=x2+1−xy = \sqrt{x^2 + 1} - x is indeed an injective (one-to-one) function. We first used the direct definition of injectivity and algebraic manipulations to show that if f(a) = f(b), then a = b. Then, we employed calculus by finding the derivative of the function and demonstrating that it is always negative, implying that the function is strictly decreasing and therefore injective. These two approaches provide a robust and comprehensive proof.

Understanding injectivity and how to prove it is a fundamental skill in mathematics. It not only helps in analyzing functions but also lays the groundwork for more advanced concepts like inverse functions and bijective mappings. I hope this detailed exploration has been helpful and has given you a clear understanding of how to tackle such problems. Keep practicing, and you'll master these concepts in no time! Remember, mathematics is all about understanding the underlying principles and applying them creatively. Keep exploring, keep questioning, and keep learning!