Examine the continuity of a function f(x) = (x - 2) (x - 3)

  1. Discontinuous at x = 2
  2. Discontinuous at x = 2, 3
  3. Continuous everywhere
  4. Discontinuous at x = 3

Answer (Detailed Solution Below)

Option 3 : Continuous everywhere
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Detailed Solution

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Concept:

  • We say f(x) is continuous at x = c if

LHL = RHL = value of f(c)

i.e., \(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{c}}^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{c}}^ + }} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{c}} \right)\)

Calculation:

\(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right)\; = \;\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} \left( {{\rm{x}} - 2} \right)\left( {{\rm{x}} - 3} \right)\)            (a ϵ Real numbers)

\(= \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{\;}}{{\rm{x}}^2} - 3{\rm{x}} - 2{\rm{x}} + 6\)

\(= \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{\;}}{{\rm{x}}^2} - 5{\rm{x}} + 6\)

\(= {{\rm{a}}^2} - 5{\rm{a}} + 6\)

∴ f(x) = f(a), So continuous at everywhere

Important tip:

Quadratic and polynomial functions are continuous at each point in their domain

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