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Question:

Find all points of discontinuity of $f$, where $f$ is defined by

$f(x) = \begin{cases} x^3 - 3 & \text{if } x \leq 2 \\ x^2 + 1 & \text{if } x > 2 \end{cases}$

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Last updated date: 20th Jun 2024
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Answer
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Hint:

To find the points of discontinuity of $f(x)$, you should examine the boundaries where the function's definition changes. For this function, the boundary is at $x = 2$. Check the limit of $f(x)$ as x approaches this point from both the left and the right.


Step-by-Step Solution:

The function $f(x)$ is defined as:

$f(x) = \begin{cases}     x^3 - 3 & \text{if } x \leq 2 \\    x^2 + 1 & \text{if } x > 2     \end{cases}   $

1. For $x \leq 2$, the function is given by $f(x) = x^3 - 3$.

Compute the function's value as x approaches 2 from the left:

$lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^-}} (x^3 - 3) = 8 - 3 = 5$


2. For $x > 2$, the function is given by $f(x) = x^2 + 1$.

Evaluate the right-hand limit as $x$ approaches 2 from the right:

$lim_{{x \to 2^+}} f(x) = \lim_{{x \to 2^+}} (x^2 + 1) = 4 + 1 = 5$


3. Compare the limits:

The left-hand limit at $x = 2$ is 5 and the right-hand limit at $x = 2$ is also 5. Thus, the two-sided limit exists and is equal to 5. Moreover, the function value at $x = 2$ is also 5 (from the left side definition). This implies that $f(x)$ is continuous at $x = 2$.


Points of Discontinuity:

The function $f(x)$ is continuous everywhere.


Note:

It's interesting to observe that even though the function has a piecewise definition, its values coincide at the point where the definition changes. This is a demonstration of how piecewise functions can be continuous across the entire domain even when they're defined by different expressions in different intervals.

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