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

Question

Discuss the continuity of the function f, where f is defined by: $f(x)=\left\{\begin{array}{ll} {2 x,} & {\text { if } x<0} \\ {0,} & {\text { if } 0 \leq x \leq 1} \\ {4 x,} & {\text { if } x>1} \end{array}\right.$

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Answer

The given function is $f(x)=\left\{\begin{array}{c} {2 x, \text { if } x \leq 0} \\ {0, \text { if } 0 \leq x \leq 1} \\ {4 x, \text { if } x>1} \end{array}\right.$
The function f is defined at all points of the real line.
Then, we have 5 cases i.e., k < 0, k = 0, 0 < k < 1, k = 1, k > 1.
Now, Case I: k < 0
Then, f(k) = 2k
$\mathop {\lim }\limits_{x \to {{k}}} f(x) = \mathop {\lim }\limits_{x \to {{k}}} (2x) = 2k = f(k)$
Thus, $\mathop {\lim }\limits_{{{x}} \to {{k}}} {{f}}({{x}}) = {{f}}({{k}})$
Hence, f is continuous at all points x such that x < 0
Case II: k = 0
f(0) = 0
$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} (2x) = 2 \times 0 = 0$
$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} (0) = 0$
$\Rightarrow \lim _{x \rightarrow \mathbf{0}^{-}} \mathrm{f}(\mathrm{x})=\lim _{x \rightarrow \mathrm{0}^{+}} \mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{k})$
Hence, f is continuous at x = 0.
Case III: 0 < k < 1
Then, f(k) = 0
$\mathop {\lim }\limits_{x \to {{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (0) = 0 = f(k)$
Thus, $\mathop {\lim }\limits_{x \to {{k}}} f(x) = f(k)$
Hence, f is continuous in (0, 1)
Case IV: k = 1
Then f(k) = f(1) = 0
$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} (0) = 0$
$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} (4x) = 4 \times 1 = 4$
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\mathbf{k}}^ - }} {\text{f}}({\text{x}}) \ne \mathop {\lim }\limits_{{\mathbf{x}} \to {{\text{k}}^ + }} {\text{f}}({\text{x}})$
Hence, f is not continuous at x = 1
Case V: k > 1
Then, f(k) = 4k
$\mathop {\lim }\limits_{x \to {{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (4x) = 4k = f(k)$
Thus, $\mathop {\lim }\limits_{{{x}} \to {{k}}} {{f}}({{x}}) = {\mathbf{f}}({{k}})$
Hence, f is continuous at all point x, s.t. x > 1
Therefore, x = 1 is the only point of discontinuity of f.