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April 1, 2015

April 1, 2015

Posted by **Catherine** on Tuesday, April 5, 2011 at 1:25am.

- Calculus -
**Damon**, Tuesday, April 5, 2011 at 2:30amf' = [4(t^2+2) -4t (2t)]/(t^2+2)^2

= [-4t^2+8]/(t^2+2)^2

0 at max or min

t = + or - sqrt 2

f" = [ (t^2+2)^2 (-8t) + 4(t^2-2)(2)(t^2+2)(2t) ] / (t^2+2)^4

if t = +sqrt 2

t^2+2 = 4

t^2-2 = 0

f" = [ -32 sqrt 2 +4*0 etc } /4^4

which is positive so a minimum

if t = - sqrt 2

t^2+2 is still 4

t^2-2 is still zero

so

f" = { +32 sqrt 2 .... etc

which is negative so a maximum

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