Posted by Bob on Thursday, January 8, 2009 at 7:36pm.
f'(x) = 2(2x)/(x^2 + 1) - 1
= 0 for max/min
for this I got x = 2 ± √5
These lie within your domain so
You will have to find
f(-3), f(2+√5)), f(2-√5) and f(5) to see which is the maximum
for f''(x) I got (4(x^2 + 1) - 4x(2x))/(x^2 + 2)^2
setting this equal to zero, I got x = ± √2
sub back in the original to find the two inflection points.
f(-3)
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