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Mathematics 17 Online
OpenStudy (cutiecomittee123):

find the relative maximum value of g(x) g(x)= 2/x^2+3

OpenStudy (dan815):

find the derivative of g(x) which is g'(x) and set to 0 solve this equation g'(x)=0

OpenStudy (cutiecomittee123):

Im still confused. Like is that all just the maximum is equal to 0?

OpenStudy (dan815):

clarification; is it... \[g(x)=\frac{ 2}{x^2+3} ~or~ g(x)=\frac{2}{x^2}+3\]

OpenStudy (dan815):

Okay well, in your case you might not even need the derivative, this question has a vertical asymptote, depending on which equation we are looking at

OpenStudy (cutiecomittee123):

Okay well how do I find the derivative?

OpenStudy (dan815):

okay but first can you tell me which one of thsoe equations is your question

OpenStudy (cutiecomittee123):

Oh my question is the fist one; (2)/(x^2+3)

OpenStudy (dan815):

okk that changes everything.. there is no vertical asymtote then, so u need to find derivative

OpenStudy (dan815):

you have to do chain rule

OpenStudy (dan815):

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