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# Question:Find all points of discontinuity of $f$, where $f$ is defined by: $f(x) = \begin{cases} |x| + 3 & \text{if } x \leq -3 \\ -2x & \text{if } -3 < x < 3 \\ 6x + 2 & \text{if } x \geq 3 \end{cases}$

Last updated date: 20th Jun 2024
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Hint:

To find the points of discontinuity for a piecewise function:

1. Identify the points where the function's definition changes.

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

3. If the left-hand limit doesn't equal the right-hand limit or the function's value, then there's a discontinuity at that point.

Step-by-step Solution:

We will analyze each piece of the function separately to determine where the discontinuities occur.

Case 1: x ≤ −3

In this case, f(x) = ∣x∣ + 3.

The graph of ∣x∣ has a sharp turn at x=0, so we might expect a discontinuity there.

However, since x ≤ −3 in this case, we can rule out any discontinuities at x = 0. The function is continuous for all x ≤ −3.

Case 2: −3 < x < 3

In this case, f(x) = −2x. This is a linear function, so it is continuous for all x in its domain, which is −3 < x < 3.

Case 3: x ≥ 3

In this case, f(x) = 6x + 2. This is another linear function, so it is continuous for all x in its domain, which is x ≥ 3.

Observations:

The only discontinuities occur at the points where the pieces of the function meet:

• $x = −3$: At this point, the one-sided limits $lim_{x\rightarrow -3^{-}} f(x)$ and $lim_{x\rightarrow -3^{+}} f(x)$ are not equal. This is because the two pieces of the function, ∣x∣+3 and −2x, have different values at x=−3. Therefore, f is discontinuous at x=−3.

• $x = 3$: At this point, the one-sided limits $lim_{x\rightarrow -3^{-}} f(x)$ and $lim_{x\rightarrow -3^{+}} f(x)$ are not equal. This is because the two pieces of the function, −2x and 6x+2, have different values at x=3. Therefore, f is discontinuous at x=3.

Therefore, the function f is discontinuous at x=−3 and x=3.

Note:

When dealing with piecewise-defined functions, always pay attention to the points where the function changes its definition. These are typically the points where discontinuities might arise. In this problem, the mismatch in function values between the second and third definitions at  $x = 3$ led to discontinuity.