Piecewise Function: Evaluation, Graph, And Continuity

by Tim Redaksi 54 views
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Hey guys! Today, we're diving deep into the world of piecewise functions. Piecewise functions might seem a bit intimidating at first, but trust me, they're super manageable once you break them down. We'll tackle evaluating them, sketching their graphs, and even checking for continuity. Let's jump right in!

Understanding Piecewise Functions

Before we get our hands dirty with the problem, let's quickly recap what piecewise functions are all about. Basically, a piecewise function is a function defined by multiple sub-functions, each applying to a certain interval of the main function's domain. Think of it like a set of rules, where each rule only applies under specific conditions. Our function for today looks like this:

f(x) = {
  x + 2, untuk x < -2
  x², untuk -2 ≤ x < 0
  2x - 3, untuk x ≥ 0
}

This tells us that if x is less than -2, we use the rule f(x) = x + 2. If x is between -2 and 0 (including -2), we use f(x) = x². And finally, if x is 0 or greater, we use f(x) = 2x - 3. Got it? Awesome, let's move on to the first part of our problem.

Evaluating the Function

Finding f(-3)

The first task is to find the value of f(-3). To do this, we need to figure out which piece of the function applies when x = -3. Since -3 is less than -2, we use the first rule: f(x) = x + 2. Plugging in -3 for x, we get:

f(-3) = -3 + 2 = -1

So, f(-3) = -1. Easy peasy, right?

Finding f(3)

Next, we need to find f(3). Now, 3 is greater than or equal to 0, so we use the third rule: f(x) = 2x - 3. Plugging in 3 for x, we have:

f(3) = 2(3) - 3 = 6 - 3 = 3

Therefore, f(3) = 3. Piece of cake!

Finding f(1)

Lastly, let's find f(1). Again, 1 is greater than or equal to 0, so we use the third rule: f(x) = 2x - 3. Plugging in 1 for x, we get:

f(1) = 2(1) - 3 = 2 - 3 = -1

So, f(1) = -1. And with that, we've nailed the first part of our problem!

Graphing the Function

Now comes the fun part: sketching the graph of our piecewise function. This might seem tricky, but we'll take it step by step. Remember, each piece of the function has its own interval where it's valid. We'll graph each piece separately and then combine them to get the complete graph.

Graphing f(x) = x + 2 for x < -2

This is a linear function with a slope of 1 and a y-intercept of 2. However, it's only valid for x < -2. So, we'll draw a line with that slope and y-intercept, but only for x values less than -2. At x = -2, the value of the function would be -2 + 2 = 0. Since the condition is x < -2, we use an open circle at the point (-2, 0) to indicate that this point is not included in the graph.

Graphing f(x) = x² for -2 ≤ x < 0

This is a quadratic function, specifically a parabola. It's only valid for x values between -2 and 0, including -2 but not including 0. Let's calculate the values at the endpoints. At x = -2, f(x) = (-2)² = 4. Since the condition is -2 ≤ x, we use a closed circle at the point (-2, 4) to indicate that this point is included in the graph. At x = 0, f(x) = (0)² = 0. Since the condition is x < 0, we use an open circle at the point (0, 0) to indicate that this point is not included in the graph. We then sketch the parabola between these two points.

Graphing f(x) = 2x - 3 for x ≥ 0

This is another linear function, with a slope of 2 and a y-intercept of -3. It's only valid for x values greater than or equal to 0. So, we'll draw a line with that slope and y-intercept, but only for x values greater than or equal to 0. At x = 0, the value of the function is 2(0) - 3 = -3. Since the condition is x ≥ 0, we use a closed circle at the point (0, -3) to indicate that this point is included in the graph.

Combining the Pieces

Now, we combine these three pieces to get the complete graph of the piecewise function. You'll see a line segment, a portion of a parabola, and another line segment, each connected (or disconnected) at the boundaries of their respective intervals. Make sure you pay close attention to the open and closed circles at the boundaries to accurately represent the function.

Checking for Continuity at x = 0

Alright, last but not least, we need to determine if our function is continuous at x = 0. Remember, for a function to be continuous at a point, three conditions must be met:

  1. f(0) must be defined.
  2. The limit of f(x) as x approaches 0 must exist.
  3. The limit of f(x) as x approaches 0 must be equal to f(0).

Let's check these conditions for our function at x = 0.

Is f(0) Defined?

Yes, f(0) is defined. According to our piecewise function, when x ≥ 0, f(x) = 2x - 3. So, f(0) = 2(0) - 3 = -3. Therefore, f(0) = -3.

Does the Limit of f(x) as x Approaches 0 Exist?

To check if the limit exists, we need to examine the left-hand limit and the right-hand limit. If they are equal, then the limit exists.

  • Left-hand limit: As x approaches 0 from the left (i.e., x < 0), we use the second piece of the function: f(x) = x². So, the left-hand limit is lim (x→0-) x² = 0² = 0.
  • Right-hand limit: As x approaches 0 from the right (i.e., x ≥ 0), we use the third piece of the function: f(x) = 2x - 3. So, the right-hand limit is lim (x→0+) (2x - 3) = 2(0) - 3 = -3.

Since the left-hand limit (0) is not equal to the right-hand limit (-3), the limit of f(x) as x approaches 0 does not exist.

Is the Limit Equal to f(0)?

Since the limit doesn't exist, we don't even need to check this condition. The function fails the second condition for continuity.

Conclusion

Because the limit of f(x) as x approaches 0 does not exist, the function is not continuous at x = 0. There you have it!

Wrapping Up

So, to recap, we successfully evaluated the piecewise function at given points, sketched its graph, and determined that it's not continuous at x = 0. Piecewise functions might seem complex at first, but by breaking them down into smaller parts, you can easily understand and work with them. Keep practicing, and you'll become a piecewise function pro in no time! Keep rocking! Let me know if you have any other questions.