Find all points of discontinuity of f , where f is defined by f ( x ) = { x 2 + x + 3 , if x 3 0 , if 0 x 1 x 2 - x + 3 , if x 1

Question

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

$f (x) =$ ${\begin{matrix} x^{2} + x + 3, & if x < 3 \\ 0, & if 0 \leq x \leq 1 \\ x^{2} - x + 3, & if x > 1 \end{matrix}$

Smart Achievers · Accepted Answer

(1)

Ans. $f$ is discontinuous at all points of $c$

Explanation:

The given function $f$ is

$f (x) =$ ${\begin{matrix} x^{2} + x + 3, & if x < 3 \\ 0, & if 0 \leq x \leq 1 \\ x^{2} - x + 3, & if x > 1 \end{matrix}$

$(i) c < 0 (ii) c > 1 (iii) 0 \leq c \leq 1$

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

Case (i): $c < 0$

then $f (x) = x^{2} + x + 3$

$f (c) = c^{2} + c + 3$

$\lim_{x \to c} f (x) = \lim_{x \to c} (x^{2} + x + 3) = c^{2} + c + 3$

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

Therefore, $f$ is continuous at all points $x$ , such that $x < 0$ .

Case (ii): $c > 1$

then $f (x) = x^{2} - x + 3$

$\lim_{x \to c} f (x) = \lim_{x \to c} (x^{2} - x + 3) = c^{2} - c + 3$

Therefore, $f$ is continuous at all points $x$ , such that $x > 1$ .

Case (iii): $0 \leq c \leq 1$

then $f (x) = 0$

$f (c) = 0$

$\lim_{x \to c} f (x) = \lim_{x \to c} 0 = 0$

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

Therefore, $f$ is continuous at all points $x$ , such that $0 \leq x \leq 1$ .

But, the given function $f$ is discontinuous at all points of $c$ .

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