Discuss the continuity of the function f, where f is defined by f(x)= 3, if 0 x 1 4, if 1<x<3 5, if 3 10

The given function is f(x)= 3, if 0 x 1 4, if 1<x<3 5, if 3 10 The function f is definedc at all points of the interval [0, 10]. Let k be the point in the interval [0, 10]. Then, we have 5 cases i. e. 0 <1, k= 1, 1<k <3, k = 3 or 3 < k 10. Now, Case I: 0 <1 Then, f(k) = 3 ^ lim _ x f(x) = ^ lim _ x (3) = 3 = f(k) Thus, ^ lim _ x f(x) = f(k) Hence, f is continuous in the interval [0, 10]. Case II: k = 1 f(1) = 3 ^ lim _ x 1^ - f(x) = ^ lim _ x 1^ - (3) = 3 ^ lim _ x 1^ + f(x) = ^ lim _ x 1^ + (4) = 4 ^ lim _ x ^ - f(x) ^ lim _ x ^ + f(x) Hence, f is not continuous at x = 1. Case III: 1 < k < 3 Then, f(x) = 4 ^ lim _ x f(x) = ^ lim _ x (4) = 4 = f(k) Thus, ^ lim _ x f(x) = f(k) Hence, f is continuous in (1, 3). Case IV: k = 3 ^ lim _ x 3^ - f(x) = ^ lim _ x 3^ - (4) = 4 ^ lim _ x 3^ + f(x) = ^ lim _ x 3^ + (5) = 5 ^ lim _ x ^ - f(x) ^ lim _ x ^ + f(x) Hence, f is not continuous at x = 3.Case V: 3 < k 10 Then, f(k) = 5 ^ lim _ x f(x) = ^ lim _ x (5) = 5 = f(k) Thus ^ lim _ x f(x) = f(k) Hence, f is contionuous at all points of the interval [3, 10]. Therefore, x = 1 and 3 are the points of discontinuity of f.

Discuss the continuity of the function f, where f is defined by f…

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Question

Discuss the continuity of the function f, where f is defined by
$\text{f(x)}= \begin{cases}\ 3,\ \ \text{if}\ 0\leq \text{x}\leq 1 \\4,\ \ \text{if}\ 1<\text{x}<3\\5,\ \text{if}\ 3\leq\text{x}\leq10\end{cases}$

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Answer

The given function is $\text{f(x)}= \begin{cases}\ 3,\ \ \text{if}\ 0\leq \text{x}\leq 1 \\4,\ \ \text{if}\ 1<\text{x}<3\\5,\ \text{if}\ 3\leq\text{x}\leq10\end{cases}$ The function f is definedc at all points of the interval [0, 10]. Let k be the point in the interval [0, 10]. Then, we have 5 cases i. e. $0\leq\text{k}<1, \text{k}= 1, 1<\text{k}$ $<3, \text{k} = 3\ \text{or}\ 3 < \text{k}\leq 10.$ Now, Case I: $0\leq\text{k}<1$ Then, f(k) = 3 $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}(3) = 3 = \text{f(k)}$ Thus, $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = \text{f(k)}$ Hence, f is continuous in the interval [0, 10]. Case II: k = 1 f(1) = 3 $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{-}}(3) = 3$ $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{+}}(4) = 4$ $\Rightarrow\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}^{-}}\text{f(x)} \neq ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}^{+}}\text{f(x)}$ Hence, f is not continuous at x = 1. Case III: 1 < k < 3 Then, f(x) = 4 $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}(4) = 4 = \text{f(k)}$ Thus, $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = \text{f(k)}$ Hence, f is continuous in (1, 3). Case IV: k = 3 $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{3}^{-}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{3}^{-}}(4) = 4$ $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{3}^{+}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{3}^{+}}(5) = 5$ $\Rightarrow\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}^{-}}\text{f(x)} \neq ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}^{+}}\text{f(x)}$ Hence, f is not continuous at x = 3.Case V: $3 < \text{k}\leq 10$
Then, f(k) = 5
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}(5) = 5 = \text{f(k)}$ Thus $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = \text{f(k)}$ Hence, f is contionuous at all points of the interval [3, 10]. Therefore, x = 1 and 3 are the points of discontinuity of f.