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find the relative maximum value of g(x) g(x)= 2/x^2+3
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find the derivative of g(x) which is g'(x) and set to 0 solve this equation g'(x)=0
Im still confused. Like is that all just the maximum is equal to 0?
clarification; is it... \[g(x)=\frac{ 2}{x^2+3} ~or~ g(x)=\frac{2}{x^2}+3\]
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
Okay well how do I find the derivative?
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okay but first can you tell me which one of thsoe equations is your question
Oh my question is the fist one; (2)/(x^2+3)
okk that changes everything.. there is no vertical asymtote then, so u need to find derivative
you have to do chain rule
|dw:1429717484042:dw|
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