Question
Asked Oct 21, 2019

find the exact max and min 5t/1+t^2

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Step 1

Step 2

...
(1+(5)-5t(2t)
f'()
(1+f)
0 5t2-105
-5=-5t2
t 1,-
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(1+(5)-5t(2t) f'() (1+f) 0 5t2-105 -5=-5t2 t 1,-

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Tagged in

Math

Calculus

Derivative