Is the function defined by f(x)= ll x+5, & if x 1 x-5, & if x>1 . a continuous function?

The given function is f(x) = l x+5, if x 1 x-5, if x>1 . The function f is defined at all points of the real line. Let k be the point on a real line. Then, we have 3 cases i.e., k 1 Now, Case I: k 1 Then, f(k) = k - 5 _ x k f(x) = _ x k (x - 5) = k - 5 Thus, _ x k f(x) = f(k) Hence, f is continuous at all real number greater than 1. Therefore, x = 1 is the only point of discontinuity of f.

Is the function defined by f(x)= ll x+5, & if x 1 x-5, & if x>1 .…

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Question

Is the function defined by
$f(x)=\left\{\begin{array}{ll} {x+5,} & {\text { if } x \leq 1} \\ {x-5,} & {\text { if } x>1} \end{array}\right.$
a continuous function?

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Answer

The given function is f(x) = $\left\{\begin{array}{l} {x+5, \text { if } x \leq 1} \\ {x-5, \text { if } x>1} \end{array}\right.$
The function f is defined at all points of the real line.
Let k be the point on a real line.
Then, we have 3 cases i.e., k < 1, or k = 1 or k > 1
Now,
Case I: k < 1
Then, f(k) = k + 5
$\mathop {\lim }\limits_{x \to {{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (x + 5) = k + 5 = f(k)$
Thus, $\mathop {\lim }\limits_{x \to k} f(x) = f(k)$
Hence, f is continuous at all real number less than 1.
Case II: k = 1
Then, f(k) = f(1) = 1 + 5 = 6
$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} (x + 5) = 1 + 5 = 6$
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(x-5)=1-5=-4$
$ \Rightarrow \mathop {\lim }\limits_{x \to {{{k}}^ - }} {\text{f}}({\text{x}}) \ne \mathop {\lim }\limits_{{\text{x}} \to {{\text{k}}^ + }} {\text{f}}({\text{x}})$
Hence, f is not continuous at x = 1.
Case III: k > 1
Then, f(k) = k - 5
$\mathop {\lim }\limits_{x \to {{k}}} f(x) = \mathop {\lim }\limits_{x \to {{k}}} (x - 5) = k - 5$
Thus, $\mathop {\lim }\limits_{x \to {{k}}} f(x) = f(k)$
Hence, f is continuous at all real number greater than 1.
Therefore, x = 1 is the only point of discontinuity of f.