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Sum
Examine the continuity of the following function :
`{:(,,f(x)= x^2 -x+9,"for",x≤3),(,,=4x+3,"for",x>3):}}"at "x=3`
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Solution
Given
`f(x)=x^2-x+9 , for x<=3`
` =4x+3 for x>3`
`f(3)=(3)^2-3+9=9-3+9`
`f(3)=15`
`Now lim_(x->3^-)f(x)=lim_(x->3)(x^2-x+9)`
` =(3)^2-(3)+9`
=15
`lim_(x->3^-)f(x)=lim_(x->3)(4x+3)`
=4(3)+3
=15
Thus from the above
`lim_(x->3^-)f(x)=lim_(x->3)f(x)=15=f(3)`
Hence function is continuous at x=3
Concept: Continuous Function of Point
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