Question

Find all points of absolute minima and maxima on the given interval.

Answer 1

\[y= -\frac{ 25 }{ t^2+5 }\]

Answer 2

What is the interval? @hdrager

Answer 3

oops sorry

Answer 4

it's [-4,2]

Answer 5

Candidates for Global Max/Min are Critical Points & Endpoints***
Find Derivative
\[y=25(t^2+5)^{-1}\]
*Power Rule/Chain Rule
\[y'=-25(t^2+5)^{-2}*(2t)\]
\[y'=\frac{ -50t }{( t^2+5)^2 }\]

Answer 6

Critical Points --> When y'=0 or y'=undefined

Critical point @ t=0
Find:
y evaluated @
t=-4
t=0
t=2

What is the biggest #? (global/absolute max)
What is the smallest #? (global/absolute min)

Answer 7

@hdrager ... any ideas?

Answer 8

wait how did you get the critical points

Answer 9

Set y' = 0 & solve for t

Answer 10

like could you go through the process of finding t?

Answer 11

\[y'=\frac{ -50t }{( t^2+5)^2 }\]

\[0=\frac{ -50t }{( t^2+5)^2 }\]

\[-50t=0\]

t=0

Remember to plug in your endpoints as well as the critical number, t=0 back into the original equation. This number will be your min/max

so
t=-4 (left bound)
t=0 (critical number)
t=2 (right bound)