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Question
Examine the continuity of f(x)=`x^2-x+9 "for" x<=3`
=`4x+3 "for" x>3, "at" x=3`
Solution
`f(x)={(x^2,-x,+9, "for" x,- <=3 ), (4x,+,3 ,"for" ,x>3):}`
`lim_(x -> 3^-)` f(x) = `lim_(x -> 3)` (x2 - x + 9)
= 9 - 3 + 9
= 15
`lim_(x -> 3^+)` f(x) = `lim_(x -> 3)` (4x + 3)
= (4 x 3) + 3
= 15 .........(i)
f(3) = 32 - 3 + 9
= 15.......(ii)
From (i) and (ii)
`lim_(x -> 3^-)` f(x) = `lim_(x -> 3^+)` f(x) = f(3)
`therefore` Function f(x) is continuous at x = 3.
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