Asked by HADEEL
Find the absolute maximum and minimum of g(x) =
(x^2 - 3)/ (x^2 + 1)
on the interval [-1; 2]:
(x^2 - 3)/ (x^2 + 1)
on the interval [-1; 2]:
Answers
Answered by
Steve
since g' = 8x/(x+1)^2
g'=0 at x=0
g'' = (8-24x^2)/(x+1)^3
g'' > 0 at x=0, so g(0) is a minimum
The maximum is attained at one or both of the endpoints.
g(-1) = -1
g(2) = 0.2
so max is attained at x=2
g'=0 at x=0
g'' = (8-24x^2)/(x+1)^3
g'' > 0 at x=0, so g(0) is a minimum
The maximum is attained at one or both of the endpoints.
g(-1) = -1
g(2) = 0.2
so max is attained at x=2
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