Problem 35 - Single Variable Calculus

Question

Determine whether the function is continuous on the closed interval. \(f(x)=\left\\{\begin{array}{ll}x+1 & \text { if } x<0 \\ 2-x & \text { if } x \geq 0\end{array}, \quad[-2,4]\right.\)

Step-by-Step Solution

Answer

The function \(f(x)=\left\\{\begin{array}{ll}x+1 & \text { if } x<0 \\\ 2-x & \text { if } x \geq 0\end{array}\right.\) is not continuous on the closed interval [-2, 4] because the left-hand limit (\(1\)) and right-hand limit (\(2\)) at x = 0 are not equal.

Step 1: Condition 1: Function value at x = 0

Using the function definition, we find that f(0) = 2 - 0 = 2.

Step 2: Condition 2: Left-hand limit at x = 0

We calculate the limit as x approaches 0 from the left (x < 0): \(\lim_{x \to 0^{-}}(x + 1) = 0 + 1 = 1\)

Step 3: Condition 3: Right-hand limit at x = 0

We calculate the limit as x approaches 0 from the right (x ≥ 0): \(\lim_{x \to 0^{+}}(2 - x) = 2 - 0 = 2\)

Step 4: Condition 4: Comparison

We compare the values from conditions 1, 2, and 3: \(f(0) = 2\), \(\lim_{x \to 0^{-}}f(x) = 1\), and \(\lim_{x \to 0^{+}}f(x) = 2\). The left-hand limit and right-hand limit are not equal at x=0. Therefore, the function is not continuous at x=0. Since the function is not continuous at x = 0, we can conclude that:

Step 5: Conclusion

The function \(f(x)=\left\\{\begin{array}{ll}x+1 & \text { if } x<0 \\\ 2-x & \text { if } x \geq 0\end{array}\right.\) is not continuous on the closed interval [-2, 4].

Key Concepts

Left-Hand Limit

When you have a piecewise function like the one given, it's crucial to check the left-hand limit. The left-hand limit describes what value the function approaches as you get closer to a certain point from the left side. For example, in our problem, at \( x = 0 \), we are interested in the value the function approaches as \( x \) comes from negative values.
For \( x < 0 \), the function is defined as \( f(x) = x + 1 \). As \( x \) approaches 0 from the left, we calculate \( \lim_{x \to 0^{-}} (x + 1) \). This limit tells us what value \( f(x) \) is getting closer to, even though \( f(x) \) might not actually equal this value.
This left-hand limit at \( x = 0 \) was found to be 1. Knowing this makes it easier to assess continuity at a specific point.

Right-Hand Limit

Now, let's consider the right-hand limit. The right-hand limit examines what happens as you approach a specific point from the right. It's like the left-hand limit but in the opposite direction.
For \( x \geq 0 \), the function is given by \( f(x) = 2 - x \). As you approach \( x = 0 \) from positive values, compute \( \lim_{x \to 0^{+}} (2 - x) \). This limit tells us the behavior of \( f(x) \) as it nears 0 from the right side.

• In our case, we found that the right-hand limit is 2.
• Comparing both this limit and the left-hand limit helps decide if the function is continuous at \( x = 0 \).

While the left-hand limit at 0 is 1, the right-hand limit is 2. This discrepancy is key in identifying discontinuities.

Function Continuity

For a function to be continuous at a specific point, three conditions need to be met:

• The function must be defined at that point, which means there is an actual value for \( f(x) \) at \( x \).
• The left-hand limit as \( x \) approaches the point must equal the right-hand limit.
• The common value of these limits should also match the function's actual value at that point.

In the given exercise, even though \( f(x) \) is defined at \( x = 0 \) (since \( f(0) = 2 \)), the left-hand limit and right-hand limit differ; they are 1 and 2, respectively.
This means \( f(x) \) is not continuous at \( x = 0 \). Due to this lack of continuity at one point within the interval, the function is not continuous on the closed interval \([-2, 4]\). Understanding these concepts of continuity can greatly help in analyzing piecewise functions accurately.