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# Question:Find all points of discontinuity of $f$, where $f$ is defined by $f(x) = \begin{cases} 2x + 3 & \text{if } x \leq 2 \\ 2x - 3 & \text{if } x > 2 \end{cases}$​

Last updated date: 09th Aug 2024
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Hint:

To determine the points of discontinuity for a piecewise function:

1. Look at the points where the function's definition changes.

2. Evaluate the function values and limits from both sides at these points.

3. If the left-hand limit does not match the right-hand limit or the function's value, the function is discontinuous at that point.

Step-by-Step Solution:

To determine the points of discontinuity of f, we need to examine its behavior at each possible point of discontinuity.

Case 1: $x < 2$

In this case, $f(x)=2x+3$.

Let's consider the one-sided limits as x approaches 2:

• Left-hand limit (LHL): $lim_{x\rightarrow 2^{-}} f(x) = lim_{x\rightarrow 2^{-}} (2x + 3) = 2(2) + 3 = 7$

• Right-hand limit (RHL): Since f(x) is not defined at x=2, the right-hand limit does not exist.

Since the LHL does not equal the RHL, f is discontinuous at x=2.

Case 2: $x > 2$

In this case, $f(x)=2x−3$.

Let's consider the one-sided limits as x approaches 2:

• Left-hand limit (LHL): Since f(x) is not defined at x=2, the left-hand limit does not exist.

• Right-hand limit (RHL): $lim_{x\rightarrow 2^{+}} + ​f(x)=lim_{x\rightarrow 2^{+}} + (2x − 3) = 2(2) − 3 = 1$

Since the LHL does not exist, we cannot determine whether f is continuous at x=2.

Therefore, the only point of discontinuity of f is $x = 2$.