Finding A, B, C Values For A Piecewise Function F(x)

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Hey everyone! Today, we're diving into a fun math problem where we need to figure out the values of a, b, and c for a piecewise function. Piecewise functions might look a little intimidating at first, but don't worry, we'll break it down step by step. This is super important because understanding piecewise functions helps us model real-world situations where different rules apply under different conditions. Think about things like tax brackets or shipping costs – they often follow a piecewise pattern.

Understanding the Piecewise Function

First, let's get clear on what a piecewise function actually is. Basically, it’s a function defined by multiple sub-functions, each applying to a certain interval of the main function's domain. So, imagine you have a function that behaves one way when x is less than 1, and another way when x is greater than or equal to 1. That’s a piecewise function in action! The key here is that we need to make sure these pieces connect smoothly, especially at the points where they transition. This often involves ensuring that the function is continuous and sometimes even differentiable at these transition points.

In our specific problem, we have a function f(x) defined as follows:

f(x) = 
  \begin{cases}
    \frac{3x^2 - 3bx + 12}{2x^2 + ax - a - 2}, & x \geq 1 \\
    \frac{2x^2 + 5x + a}{x^3 - 1}, & x < 1 \\
    2 + c, & x = 1
  \end{cases}

This looks a bit complex, but let's break it down. We've got three parts here:

  1. For x greater than or equal to 1, f(x) is defined by the rational function (3x^2 - 3bx + 12) / (2x^2 + ax - a - 2).
  2. For x less than 1, f(x) is defined by another rational function, (2x^2 + 5x + a) / (x^3 - 1).
  3. And at the specific point x = 1, f(x) is equal to 2 + c.

Our mission, should we choose to accept it (and we do!), is to find the values of a, b, and c that make this function tick. The most important concept we'll use is continuity. For a piecewise function to be truly well-behaved, the different pieces need to meet up nicely at the boundaries. This means the left-hand limit and the right-hand limit at x = 1 must be equal, and they must also equal the function's value at x = 1. So, let's dive into how we can use this idea to solve for our unknowns.

Ensuring Continuity at x = 1

Okay, so the big idea here is continuity. For our piecewise function to be continuous at x = 1, three things need to happen:

  1. The limit of f(x) as x approaches 1 from the left (i.e., x < 1) must exist.
  2. The limit of f(x) as x approaches 1 from the right (i.e., x ≥ 1) must exist.
  3. These two limits must be equal, and they must also be equal to the function's value at x = 1, which is given as f(1) = 2 + c.

Let's tackle these conditions one by one.

1. Limit from the Left (x < 1)

When x is less than 1, we use the sub-function:

f(x) = \frac{2x^2 + 5x + a}{x^3 - 1}

To find the limit as x approaches 1 from the left, we write:

\lim_{x \to 1^-} \frac{2x^2 + 5x + a}{x^3 - 1}

Now, if we directly substitute x = 1, we get:

\frac{2(1)^2 + 5(1) + a}{(1)^3 - 1} = \frac{7 + a}{0}

Uh oh! We've got a division by zero situation, which is a big no-no in the math world. For this limit to exist (and be a finite number), the numerator must also be zero when x = 1. This is because we can then try to factor and simplify the expression, hopefully getting rid of the troublesome (x - 1) term in the denominator. So, we need:

7 + a = 0

This gives us our first crucial piece of the puzzle:

a = -7

Awesome! Now we know the value of a. Let's substitute this back into our expression and see if we can simplify it. Our function for x < 1 now looks like:

f(x) = \frac{2x^2 + 5x - 7}{x^3 - 1}

We know that both the numerator and denominator must have a factor of (x - 1) since they both evaluate to zero at x = 1. Let's factor them. Factoring the numerator 2x^2 + 5x - 7 gives us (x - 1)(2x + 7). Factoring the denominator x^3 - 1 (which is a difference of cubes) gives us (x - 1)(x^2 + x + 1). So, our expression becomes:

f(x) = \frac{(x - 1)(2x + 7)}{(x - 1)(x^2 + x + 1)}

Now we can cancel out the (x - 1) terms (since we're looking at the limit as x approaches 1, not actually at x = 1), and we get:

f(x) = \frac{2x + 7}{x^2 + x + 1}

Now, let's find the limit as x approaches 1 from the left:

\lim_{x \to 1^-} \frac{2x + 7}{x^2 + x + 1} = \frac{2(1) + 7}{(1)^2 + 1 + 1} = \frac{9}{3} = 3

So, the limit from the left is 3. Great job! We're one step closer to solving this puzzle.

2. Limit from the Right (x ≥ 1)

Now let's tackle the limit as x approaches 1 from the right. For x ≥ 1, our function is:

f(x) = \frac{3x^2 - 3bx + 12}{2x^2 + ax - a - 2}

We already found that a = -7, so let's substitute that in:

f(x) = \frac{3x^2 - 3bx + 12}{2x^2 - 7x + 7 - 2} = \frac{3x^2 - 3bx + 12}{2x^2 - 7x + 5}

Now, let's find the limit as x approaches 1 from the right:

\lim_{x \to 1^+} \frac{3x^2 - 3bx + 12}{2x^2 - 7x + 5}

If we try direct substitution again, we get:

\frac{3(1)^2 - 3b(1) + 12}{2(1)^2 - 7(1) + 5} = \frac{15 - 3b}{0}

Another division by zero! Just like before, this means the numerator must also be zero when x = 1 for the limit to exist. So, we need:

15 - 3b = 0

Solving for b, we get:

3b = 15
b = 5

Fantastic! We've found b = 5. Now, let's plug that back into our function:

f(x) = \frac{3x^2 - 15x + 12}{2x^2 - 7x + 5}

We know both the numerator and denominator must have a factor of (x - 1). Let's factor them. The numerator 3x^2 - 15x + 12 factors to 3(x - 1)(x - 4). The denominator 2x^2 - 7x + 5 factors to (x - 1)(2x - 5). So, we have:

f(x) = \frac{3(x - 1)(x - 4)}{(x - 1)(2x - 5)}

Cancel out the (x - 1) terms:

f(x) = \frac{3(x - 4)}{2x - 5}

Now, let's find the limit as x approaches 1 from the right:

\lim_{x \to 1^+} \frac{3(x - 4)}{2x - 5} = \frac{3(1 - 4)}{2(1) - 5} = \frac{3(-3)}{-3} = 3

The limit from the right is also 3! This is great news because it matches the limit from the left, which we found earlier. We're on the right track.

3. Function Value at x = 1

We know that f(1) = 2 + c. For continuity, this value must be equal to the limits we just calculated, which are both 3. So:

2 + c = 3

Solving for c, we get:

c = 1

And there you have it! We've found all the values: a = -7, b = 5, and c = 1.

Final Solution and Verification

So, to recap, we've determined the values of a, b, and c that make our piecewise function continuous at x = 1. Here's what we found:

  • a = -7
  • b = 5
  • c = 1

Let's plug these values back into our original piecewise function to verify our solution:

f(x) = 
  \begin{cases}
    \frac{3x^2 - 15x + 12}{2x^2 - 7x + 5}, & x \geq 1 \\
    \frac{2x^2 + 5x - 7}{x^3 - 1}, & x < 1 \\
    3, & x = 1
  \end{cases}

We already showed that the limits from both sides are 3. And indeed, f(1) = 2 + c = 2 + 1 = 3. So, everything checks out!

Importance of Continuity

Continuity is a huge deal in calculus and many areas of math and physics. A continuous function is one that doesn't have any sudden jumps or breaks. You can think of it as a function whose graph you can draw without lifting your pencil from the paper. In practical terms, continuity often represents the smooth transition between different states or conditions. For instance, in physics, we might want to model the continuous motion of an object, and discontinuities would represent instantaneous jumps in position or velocity, which are usually not physically realistic.

In our piecewise function example, ensuring continuity meant that the different parts of the function