Find all points of discontinuity of $f$,where $f$ is defined by $f(x) = \begin{cases} x^{10} - 1, & \text{if } x \le 1 \\ x^2, & \text{if } x > 1 \end{cases}$.
(A) The given function is $f(x) = \begin{cases} x^{10} - 1, & \text{if } x \le 1 \\ x^2, & \text{if } x > 1 \end{cases}$.
The function $f$ is defined for all real numbers.
Case $I$: If $c < 1$,then $f(c) = c^{10} - 1$. The limit $\lim_{x \to c} f(x) = \lim_{x \to c} (x^{10} - 1) = c^{10} - 1 = f(c)$. Thus,$f$ is continuous for all $x < 1$.
Case $II$: If $c = 1$,we check the limits at $x = 1$.
The left-hand limit is $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^{10} - 1) = 1^{10} - 1 = 0$.
The right-hand limit is $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2) = 1^2 = 1$.
Since the left-hand limit $(0)$ is not equal to the right-hand limit $(1)$,the function $f$ is discontinuous at $x = 1$.
Case $III$: If $c > 1$,then $f(c) = c^2$. The limit $\lim_{x \to c} f(x) = \lim_{x \to c} (x^2) = c^2 = f(c)$. Thus,$f$ is continuous for all $x > 1$.
Conclusion: The only point of discontinuity is $x = 1$.
The function $f$ is defined for all real numbers.
Case $I$: If $c < 1$,then $f(c) = c^{10} - 1$. The limit $\lim_{x \to c} f(x) = \lim_{x \to c} (x^{10} - 1) = c^{10} - 1 = f(c)$. Thus,$f$ is continuous for all $x < 1$.
Case $II$: If $c = 1$,we check the limits at $x = 1$.
The left-hand limit is $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^{10} - 1) = 1^{10} - 1 = 0$.
The right-hand limit is $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2) = 1^2 = 1$.
Since the left-hand limit $(0)$ is not equal to the right-hand limit $(1)$,the function $f$ is discontinuous at $x = 1$.
Case $III$: If $c > 1$,then $f(c) = c^2$. The limit $\lim_{x \to c} f(x) = \lim_{x \to c} (x^2) = c^2 = f(c)$. Thus,$f$ is continuous for all $x > 1$.
Conclusion: The only point of discontinuity is $x = 1$.
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