#### Question

Examine the continuity of the following function :

`f(x)=x^2-x+9` , for x ≤ 3

`=4x+3` for 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

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#### Reference Material

Solution for question: Examine the Continuity of the Following Function concept: Continuous Function of Point. For the courses HSC Commerce (Marketing and Salesmanship), HSC Commerce