Discontinuity Points: A Piecewise Function Example

Hey guys! Today, we're diving into a fun little problem about finding where a piecewise function decides to be a bit of a rebel and become discontinuous. Piecewise functions can sometimes be tricky because they're defined differently over different intervals, and it's at the boundaries of these intervals where things can get interesting – and potentially discontinuous!

Understanding Discontinuity

Before we jump into the problem, let's make sure we're all on the same page about what discontinuity actually means. In simple terms, a function is discontinuous at a point if there's a break, jump, or hole in the graph at that point. More formally, for a function f(x) to be continuous at a point x = a, three conditions must be met:

1. f(a) must be defined (i.e., the function has a value at x = a).
2. The limit of f(x) as x approaches a must exist (i.e., the left-hand limit and the right-hand limit must be equal).
3. The limit of f(x) as x approaches a must be equal to f(a).

If any of these conditions are not met, then the function is discontinuous at x = a. So, we need to check at the boundaries.

The Piecewise Function

Here's the function we're going to analyze:

$f(x)=\left\{\begin{array}{ll} 6 x+1 & \text { if } x<-2 \\ 5 x-1 & \text { if }-2 \leq x<4 \\ -6 x^2+115 & \text { if } x \geq 4 \end{array}\right.$

This function is defined in three parts:

• For x < -2, f(x) = 6x + 1
• For -2 ≤ x < 4, f(x) = 5x - 1
• For x ≥ 4, f(x) = -6x² + 115

The points where the function might be discontinuous are the points where the definition changes, which are x = -2 and x = 4. We need to investigate these points.

Checking for Discontinuity at x = -2

Okay, let's start by checking x = -2. We need to check the three conditions for continuity mentioned earlier. First, let's evaluate the function at x = -2. Since x = -2 falls into the second interval, we use the second definition of the function:

f(-2) = 5(-2) - 1 = -10 - 1 = -11

So, f(-2) = -11 is defined.

Next, we need to find the left-hand limit and the right-hand limit as x approaches -2.

• Left-hand limit: As x approaches -2 from the left (x < -2), we use the first definition of the function:

  lim x→-2- f(x) = lim x→-2- (6x + 1) = 6(-2) + 1 = -12 + 1 = -11

• Right-hand limit: As x approaches -2 from the right (x ≥ -2), we use the second definition of the function:

  lim x→-2+ f(x) = lim x→-2+ (5x - 1) = 5(-2) - 1 = -10 - 1 = -11

Since the left-hand limit and the right-hand limit are both equal to -11, the limit exists and is equal to -11. Also, f(-2) = -11, which is equal to the limit. Therefore, all three conditions for continuity are satisfied at x = -2, so the function is continuous at x = -2. Phew, one down!

Checking for Discontinuity at x = 4

Now, let's move on to x = 4. Again, we need to check the three conditions for continuity. First, let's evaluate the function at x = 4. Since x = 4 falls into the third interval, we use the third definition of the function:

f(4) = -6(4)² + 115 = -6(16) + 115 = -96 + 115 = 19

So, f(4) = 19 is defined.

Next, we need to find the left-hand limit and the right-hand limit as x approaches 4.

• Left-hand limit: As x approaches 4 from the left (x < 4), we use the second definition of the function:

  lim x→4- f(x) = lim x→4- (5x - 1) = 5(4) - 1 = 20 - 1 = 19

• Right-hand limit: As x approaches 4 from the right (x ≥ 4), we use the third definition of the function:

  lim x→4+ f(x) = lim x→4+ (-6x² + 115) = -6(4)² + 115 = -6(16) + 115 = -96 + 115 = 19

Since the left-hand limit and the right-hand limit are both equal to 19, the limit exists and is equal to 19. Also, f(4) = 19, which is equal to the limit. Therefore, all three conditions for continuity are satisfied at x = 4, so the function is continuous at x = 4. Another one bites the dust!

Conclusion

After checking both potential points of discontinuity, x = -2 and x = 4, we found that the function is continuous at both points. Therefore, the function f(x) is continuous everywhere, and there are no points of discontinuity.

So, there you have it! Piecewise functions might look intimidating, but by carefully checking the conditions for continuity at the boundaries of each piece, you can determine whether they are continuous or not. Remember to evaluate the function at the point, find the left-hand and right-hand limits, and check if they all match up. If they do, you've got yourself a continuous function at that point. If not, you've found a point of discontinuity. Keep practicing, and you'll become a pro at spotting discontinuities in no time!

Key Points:

• Discontinuity occurs where a function has breaks, jumps, or holes.
• Piecewise functions are defined differently over different intervals.
• To check for continuity, evaluate the function and find the left-hand and right-hand limits at the boundary points.
• If the function value and the limits all match, the function is continuous at that point.

Happy calculating, folks! I hope this was helpful, and remember, math can be fun if you approach it step by step!