Find all points of discontinuous of f , where f is defined by f ( x ) = { x + 3 , if x 3 x - 3 , if x 3

Question

Find all points of discontinuous of $f$ , where $f$ is defined by

$f (x) =$ ${\begin{matrix} x + 3, & if x \leq 3 \\ x - 3, & if x > 3 \end{matrix}$

Smart Achievers · Accepted Answer

(1)

Ans. $f$ is discontinuous at $x = 3$

Explanation:

The given function $f$ is

$f (x) =$ ${\begin{matrix} x + 3, & if x \leq 3 \\ x - 3, & if x > 3 \end{matrix}$

$(i) c < 3 (ii) c = 3 (iii) c > 3$

where $c$ be a point on the real line. Then these three cases are valid.

Case (i): $c < 3$

Then, $f (x) = x + 3$

$f (c) = c + 3$

$\lim_{x \to c} f (x) = \lim_{x \to c} (x + 3) = c + 3$

$∴$ $\lim_{x \to c} f (x) = f (c)$

Therefore, $f$ is continuous at all points $x$ , such that $x < 3$

Case (ii): $c = 3$

Then, $f (x) = x + 3$

$\lim_{x \to c^{-}} f (x) = \lim_{x \to 3^{-}} (x + 3) = 3 + 3 = 6$

$\lim_{x \to c^{+}} f (x) = \lim_{x \to 3^{+}} (x - 3) = 3 - 3 = 0$

The value of the function $f$ at $c = 3$ is $f (3) = 3 + 3 = 6$

Since $\lim_{x \to 3^{-}} f (x) \neq \lim_{x \to 3^{+}} f (x),$ the limit does not exist.

Hence, $f$ is discontinuous at $x = 3$ .

Case (iii): $c > 3$

Then, $f (x) = x - 3$

$f (c) = c - 3$

$\lim_{x \to c} f (x) = \lim_{x \to c} (x - 3) = c - 3$

$∴$ $\lim_{x \to c} f (x) = f (c)$

Therefore, $f$ is continuous at all points $x$ , such that $x > 3$ .

The given function $f$ is discontinuous at $x = 3$ .

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