Continuity: Finding 'a' In A Piecewise Function

Hey guys, let's dive into a cool math problem today involving piecewise functions and continuity. We're going to tackle how to find the value of a constant, 'a', that makes a function continuous. This is a super common type of problem in calculus, and once you get the hang of it, you'll be solving them in no time! So, buckle up, and let's get started on understanding how to determine 'a' for a function to be continuous.

Understanding the Function and Continuity

First off, let's break down the function we're working with. We have a function f(x) defined on all real numbers (ℝ) in two parts. The point where the definition changes is at x = -1. For values of x less than -1, the function is defined as f(x) = √(x² + 3) + a·x + 1. This part involves a square root and a linear term with our constant 'a'. For values of x greater than or equal to -1, the function is defined as f(x) = (x² + x + 1)/(x + 2). This is a rational function. Our goal, specifically in question 1, is to determine 'a' for that f is continuous over its entire domain. The curve representing this function is called ζ (zeta).

What Does Continuity Mean, Anyway?

Before we jump into solving for 'a', let's quickly recap what it means for a function to be continuous. Imagine drawing the graph of a function without lifting your pen from the paper. If you can do that, the function is continuous. Mathematically, a function f(x) is continuous at a point c if three conditions are met:

1. f(c) is defined: The function must have a value at that point.
2. The limit of f(x) as x approaches c exists: This means the function approaches the same value from both the left and the right side of c.
3. The limit of f(x) as x approaches c equals f(c): The value the function approaches must be the actual value of the function at c.

When we talk about a function being continuous over its entire domain, it means it's continuous at every single point. For piecewise functions like ours, the critical point to check is where the definition changes – in this case, at x = -1. We also need to ensure that each piece is continuous on its own interval. The first piece, √(x² + 3) + a·x + 1, is continuous for all x < -1 because square roots of positive numbers and linear functions are generally continuous. Similarly, the second piece, (x² + x + 1)/(x + 2), is continuous for all x ≥ -1 except where the denominator is zero. The denominator x + 2 is zero when x = -2. Since our second piece is only defined for x ≥ -1, the point x = -2 is not within this interval, so this rational function is continuous on its domain x ≥ -1.

Therefore, the only point we really need to focus on to make the entire function continuous is the transition point, x = -1. To ensure f is continuous at x = -1, the function must approach the same value from the left side of -1 as it does from the right side of -1, and this value must be equal to the function's value at -1.

Applying the Continuity Conditions at x = -1

Alright, let's put our math hats on and apply these conditions to our specific function at x = -1. We need to make sure the limit from the left equals the limit from the right, and both equal the function's value at x = -1.

1. The Limit as x approaches -1 from the left (x < -1)

When x is approaching -1 from values less than -1, we use the first definition of f(x): f(x) = √(x² + 3) + a·x + 1. So, we need to find the limit:

$$\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (\sqrt{x^2 + 3} + a \cdot x + 1)$$

Since this expression is continuous, we can simply substitute x = -1 into it:

$$\sqrt{(-1)^2 + 3} + a \cdot (-1) + 1 = \sqrt{1 + 3} - a + 1 = \sqrt{4} - a + 1 = 2 - a + 1 = 3 - a$$

So, the limit from the left is 3 - a.

2. The Limit as x approaches -1 from the right (x ≥ -1)

When x is approaching -1 from values greater than or equal to -1, we use the second definition of f(x): f(x) = (x² + x + 1)/(x + 2). We need to find the limit:

$$\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} \frac{x^2 + x + 1}{x + 2}$$

Again, this rational function is continuous at x = -1 (since the denominator isn't zero), so we can substitute x = -1:

$$\frac{(-1)^2 + (-1) + 1}{(-1) + 2} = \frac{1 - 1 + 1}{1} = \frac{1}{1} = 1$$

So, the limit from the right is 1.

3. The Value of the Function at x = -1 (x ≥ -1)

For the value of the function at x = -1, we also use the second definition because it includes x = -1 (x ≥ -1):

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

As expected, since the second piece is continuous at x = -1, the limit from the right is equal to the function's value at x = -1. This gives us a value of 1.

Solving for 'a'

Now, for the function f to be continuous at x = -1, the limit from the left must equal the limit from the right (and the function value at x = -1). We already found:

• Limit from the left: 3 - a
• Limit from the right (and f(-1)): 1

So, we set these equal to each other to determine 'a' for that f is continuous:

$$3 - a = 1$$

Now, it's just a simple algebraic solve for 'a':

$$3 - 1 = a$$

$$a = 2$$

Boom! We've found it. By setting the limits from both sides equal, we've determined that a = 2 makes the function f(x) continuous at the crucial point x = -1. This means that if you were to graph this function with a = 2, you could draw it without lifting your pen at x = -1.

Conclusion

So there you have it, guys! Finding the value of 'a' to ensure continuity in a piecewise function boils down to understanding the definition of continuity and carefully applying it at the point where the function's definition changes. We calculated the limit from the left, the limit from the right, and the function's value at the transition point (x = -1). By setting the appropriate limits equal, we were able to solve for 'a'. In this case, a = 2 is the magic number that makes our function f(x) continuous everywhere. Keep practicing these types of problems, and you'll become a continuity pro in no time! Let me know if you have any questions in the comments below!