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# Question:Find all points of discontinuity of $f$, where $f$ is defined by $f(x) = \begin{cases} x^{10} - 1 & \text{if } x \leq 1 \\x^2 & \text{if } x > 1 \end{cases}$

Last updated date: 06th Sep 2024
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Answer
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Hint:

To identify points of discontinuity, focus on the boundaries where the function's definition changes. In this case, the critical boundary is at $x = 1$. Compute the limits of $f(x)$ as $x$ approaches this point from both the left and the right.

Step-by-Step Solution:

The function $f(x)$ is given by:

$f(x) = \begin{cases} x^{10} - 1 & \text{if } x \leq 1 \\x^2 & \text{if } x > 1 \end{cases}$

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1. For $x \leq 1$, the function is $f(x) = x^{10} - 1$. Evaluate the function's value as $x$ approaches 1 from the left:

$lim_{{x \to 1^-}} f(x) = \\lim_{{x \to 1^-}} (x^{10} - 1) = 1^{10} - 1 = 0$

2. For $x > 1$, the function is $( f(x) = x^2$ . Compute the right-hand limit as $( x$ approaches 1 from the right:

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

3. Compare the limits:

The left-hand limit at $x = 1$ is 0 and the right-hand limit at $x = 1$ is 1. Since the left-hand limit does not equal the right-hand limit, the function is discontinuous at $x = 1$.

Points of Discontinuity:

The function $f(x)$ is discontinuous at $x = 1$.

Note:

Discontinuities often occur at the boundaries of piecewise functions, especially if the functional forms on either side of the boundary do not match up. In this case, the power of 10 in the polynomial for $x \leq 1$ and the quadratic function for $x > 1$ resulted in a jump discontinuity at $x = 1$.