Question
Is the function defined by $\text{f(x)}= \begin{cases}\text{x} + 5,\ \ \text{if x}\leq 1 \\\text{x} - 5,\ \ \text{if x}>1\end{cases}$
a continuous function?
a continuous function?
f(c) = c
+ 5 $\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = \text{f(c)}$ $\therefore$ f is continuous at all points x < 1. Case II: c > 1 $^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x} - 5) = \text{c} - 5$f(c) = c
- 5 $\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = \text{f(c)}$ $\therefore$ f is continuous at all points x > 1. Case III: c = 1 $^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{(x} + 5) = 1 + 5 = 6$ $^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{(x} - 5) = 1 - 5 = -4$ $\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)} \neq ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)}$ $\therefore$ f is discontinuous at x = 1.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.