Question

Which one of the following functions is continuous at x = 3?
1. \(f\left( x \right)\left\{ {\begin{array}{*{20}{c}} {2,\;\;\;\;if\;\;\;x = 3}\\ {x - 1,\;\;\;\;if\;\;\;x > 3}\\ {\frac{{x + 3}}{3},\;\;\;\;if\;\;\;\;x < 3} \end{array}} \right.\)
2. \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {4,\;\;\;\;if\;\;\;x = 3}\\ {8 - x\;\;\;if\;\;\;x \ne 3} \end{array}} \right.\)
3. \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {x + 3,\;\;\;\;if\;\;\;x \le 3}\\ {x - 4\;\;\;\;if\;\;\;x > 3} \end{array}} \right.\)
4. \(f\left( x \right) = \frac{1}{{{x^3} - 27}}\:if\:x\ne3\)

Answer 1

Correct Answer - Option 1 : \(f\left( x \right)\left\{ {\begin{array}{*{20}{c}} {2,\;\;\;\;if\;\;\;x = 3}\\ {x - 1,\;\;\;\;if\;\;\;x > 3}\\ {\frac{{x + 3}}{3},\;\;\;\;if\;\;\;\;x < 3} \end{array}} \right.\)

The correct answer is option 1.

If \(Lt_{x \to a^-} f(x) = Lt_{ x \to a^+ } f(x)=f(a) \space then f(x) is \space continuous \space at \space x=a \)

Option 1: Correct

\(f\left( x \right)\left\{ {\begin{array}{*{20}{c}} {2,\;\;\;\;if\;\;\;x = 3}\\ {x - 1,\;\;\;\;if\;\;\;x > 3}\\ {\frac{{x + 3}}{3},\;\;\;\;if\;\;\;\;x < 3} \end{array}} \right.\)

\(f(3) = 2\)

\(\\Lt_{x \to 3^+} f(x) = Lt_{x \to 3} {(x-1)}=2, \)

\(Lt_{x \to 3^-} f(x) = Lt_{x \to 3} {(x+3) \over 3}=2,\)

 

Therefore the function is continuous at x = 3