Question

Find the absolute maximum and absolute minimum values of on the given interval. f(t)= t radical 4-t^2 [-1,2]

Answer 1

ah now we have a question

Answer 2

take the derivative, check the critical points and the endpoints.

Answer 3

im having trouble wit the radical

Answer 4

derivative is found using product rule

Answer 5

i got to do product rule and chain rule right

Answer 6

you get
\[f'(t)=\sqrt{4-t^2}+\frac{t}{\sqrt{4-t^2}}\times -2t\]

Answer 7

right product plus chain. then add

Answer 8

okay i only got up to the derivative but didnt no how to combine

Answer 9

you get
\[\frac{2(2-t^2)}{\sqrt{4-t^2}}\]

Answer 10

and then just use the numerator equal to zero?

Answer 11

oh you have to multiply first one top and bottom by
\[\sqrt{4-t^2}\]

Answer 12

so you get
\[\frac{4-t^2}{\sqrt{4-t^2}}-\frac{2t^2}{\sqrt{4-t^2}}\]] etc

Answer 13

okay i get it after that

Answer 14

wait i messed up sorry

Answer 15

i forgot the 2 in the denominator of the second one. it should be
\[\frac{4-t^2}{\sqrt{4-t^2}}-\frac{t^2}{\sqrt{4-t^2}}\]

Answer 16

so numerator is
\[4-2t^2=2(2-t^2)\] zero at
\[\sqrt{2}\]

Answer 17

also
\[-\sqrt{2}\] but that is not in your interval

Answer 18

okay im good after that

Answer 19

ok don't forget to evaluate at the endpoints

Answer 20

just had trouble with the critical point

Answer 21

thank you

Answer 22

yw