Continuity Of A Piecewise Function: Finding A & B Values

Hey guys! Today, we're diving into the fascinating world of piecewise functions and exploring how to ensure they're continuous. We've got a fun problem on our hands: figuring out the values of real numbers a and b that make a given piecewise function continuous. This is a crucial concept in calculus, and mastering it will definitely level up your math game. So, let’s break down the problem step by step and make sure we understand every detail. By the end of this, you'll be a pro at tackling similar continuity challenges! Remember, the key is understanding the conditions for continuity and applying them systematically. Let’s jump right in!

The Function at Hand

Okay, so we're given a function f(x) that's defined in two parts. For those of you who might be new to this, a piecewise function is basically a function that has different rules for different intervals of its input. It’s like having two or more functions stitched together to make a single function. In our case, we have:

${f(x) = \begin{cases} \frac{x^2-1}{ax-b^2} & \text{if } x \leq 0 \\ \frac{3x+a}{x^2+b} & \text{if } x > 0 \end{cases}}$

This looks a bit intimidating at first glance, but don't worry, we'll break it down. The top part, (x^2 - 1) / (ax - b^2), applies when x is less than or equal to 0. The bottom part, (3x + a) / (x^2 + b), applies when x is greater than 0. Our mission, should we choose to accept it (and we do!), is to find the values of a and b that make this function continuous everywhere. This means there are no sudden jumps or breaks in the graph of the function.

Understanding Continuity: The Key to Success

Before we dive into the algebra, let's quickly recap what it means for a function to be continuous. A function f(x) is continuous at a point x = c if three conditions are met:

1. f(c) is defined (i.e., the function has a value at c).
2. The limit of f(x) as x approaches c exists (i.e., the function approaches a specific value as we get closer to c from both sides).
3. The limit of f(x) as x approaches c is equal to f(c) (i.e., the value the function approaches is the actual value of the function at c).

For a piecewise function like ours to be continuous, it needs to be continuous on each interval and, most importantly, at the point where the pieces connect. In our case, that critical point is x = 0. This is where we’ll focus our attention, ensuring that the left-hand limit, the right-hand limit, and the function's value all match up. Remember, the goal is to make sure there are no gaps or jumps at this connection point. So, let’s put on our detective hats and get to work!

Analyzing Continuity at x = 0

Alright, let's zoom in on the point where things get interesting: x = 0. This is the junction where our two pieces of the function meet, and it's crucial that they meet smoothly for the function to be continuous. To ensure continuity at x = 0, we need to check three things, just like we discussed:

1. The function must be defined at x = 0.
2. The limit of the function as x approaches 0 must exist.
3. The function's value at 0 must equal the limit as x approaches 0.

Let’s tackle these one by one. First, let’s look at the function's value at x = 0. Since the first piece of our function, (x^2 - 1) / (ax - b^2), is defined for x ≤ 0, we'll use that to find f(0). Plugging in x = 0, we get:

${f(0) = \frac{0^2 - 1}{a(0) - b^2} = \frac{-1}{-b^2} = \frac{1}{b^2}}$

So, f(0) is 1/b^2, but there's a catch! This is only defined if b is not equal to 0. If b were 0, we'd be dividing by zero, which is a big no-no in the math world. So, we've already learned something important: b cannot be 0. This is our first constraint.

Examining the Limits: Left and Right

Now, let’s move on to the second condition: the limit as x approaches 0. For the limit to exist at x = 0, the left-hand limit (as x approaches 0 from the left) and the right-hand limit (as x approaches 0 from the right) must both exist and be equal. This is like making sure two roads meet at the same point – if they don’t, there's a discontinuity.

Let's start with the left-hand limit. As x approaches 0 from the left (x < 0), we use the first piece of our function:

${\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x^2 - 1}{ax - b^2}}$

We can plug in x = 0 directly into this expression (as long as the denominator isn't zero, which we've already addressed), and we get:

${\lim_{x \to 0^-} f(x) = \frac{0^2 - 1}{a(0) - b^2} = \frac{-1}{-b^2} = \frac{1}{b^2}}$

Great! The left-hand limit is 1/b^2. Now, let’s tackle the right-hand limit. As x approaches 0 from the right (x > 0), we use the second piece of our function:

${\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{3x + a}{x^2 + b}}$

Again, we can plug in x = 0 directly, and we get:

${\lim_{x \to 0^+} f(x) = \frac{3(0) + a}{0^2 + b} = \frac{a}{b}}$

So, the right-hand limit is a/b. For the overall limit to exist at x = 0, these two limits must be equal. This gives us our second important equation:

${\frac{1}{b^2} = \frac{a}{b}}$

This is a crucial relationship between a and b that we’ll need to solve. We're getting closer to cracking this problem wide open!

Solving for a and b: Putting the Pieces Together

Okay, we've got some key pieces of the puzzle laid out. We know that for our piecewise function to be continuous at x = 0, two main conditions must be met:

1. b cannot be 0 (we established this when looking at f(0)).
2. The left-hand limit and the right-hand limit at x = 0 must be equal, which gave us the equation 1/b^2 = a/b.

Now, let's use these conditions to actually solve for a and b. We’ll start with the equation we derived from the limits:

${\frac{1}{b^2} = \frac{a}{b}}$

To make this equation a bit easier to work with, we can multiply both sides by b^2. Remember, we know b isn't 0, so we're not multiplying by zero, which would mess things up. Multiplying both sides by b^2 gives us:

${1 = ab}$

This is a much simpler equation! It tells us that the product of a and b must be 1. This is a crucial relationship that narrows down the possible values of a and b significantly. It implies that a and b are reciprocals of each other; in other words, a = 1/b and b = 1/a.

The Third Condition: Ensuring the Limit Equals the Function Value

We've handled the first two conditions for continuity, but there's one more to tackle. We need to ensure that the limit of f(x) as x approaches 0 is equal to the function's value at x = 0. We already know:

• f(0) = 1/b^2
• The limit as x approaches 0 (from both sides) is 1/b^2 (from the left-hand limit) and a/b (from the right-hand limit).

We've already ensured that the left-hand limit equals the right-hand limit by setting 1/b^2 = a/b. Now, we just need to make sure this common limit value matches f(0), which it already does! So, this third condition doesn't give us any new constraints in this case, but it’s always good to check to be thorough.

Finalizing the Solution: Finding the Values

So, where do we stand? We've determined that for the piecewise function to be continuous, the following must be true:

• b ≠ 0
• ab = 1

These two conditions are the key to our solution. The equation ab = 1 is the most important result. It tells us that for any non-zero value of b, we can find a corresponding value of a that makes the function continuous. Specifically, a = 1/b. This means we have infinitely many solutions! For every real number b (except 0), there's a value of a (which is 1/b) that makes our function continuous.

To illustrate, let's pick a few values for b and see what a would be:

• If b = 1, then a = 1/1 = 1
• If b = 2, then a = 1/2
• If b = -1, then a = 1/(-1) = -1
• If b = 0.5, then a = 1/0.5 = 2

And so on! We can see that there are countless pairs of (a, b) that satisfy our condition. This is a fantastic result because it shows how a single equation (ab = 1) can define a whole set of solutions.

Conclusion: Tying It All Together

Wow, we've really taken this problem apart and put it back together! Let's recap what we've accomplished. We started with a piecewise function and a mission to find the values of a and b that make it continuous. To do this, we:

1. Reviewed the definition of continuity at a point.
2. Focused on the critical point x = 0 where the pieces of the function meet.
3. Calculated the left-hand limit and the right-hand limit at x = 0.
4. Set these limits equal to each other to ensure the overall limit exists.
5. Ensured that the limit at x = 0 equals the function's value at x = 0.
6. Solved the resulting equation to find the relationship between a and b.

Our final answer is that the function is continuous for all real values of b (except 0), where a = 1/b. This is a beautiful example of how the abstract concepts of limits and continuity can lead to concrete solutions and a deeper understanding of functions.

So, guys, the next time you encounter a piecewise function and need to check for continuity, remember the steps we've discussed here. Break it down, analyze the limits, and solve the equations. You've got this! And remember, math isn't just about numbers and equations; it's about problem-solving and critical thinking, skills that will serve you well in all aspects of life. Keep practicing, keep exploring, and keep having fun with math!