Question

I'm stuck: Finding critical numbers

Answer 1

\[f(t)= t \sqrt{4-t^2}, [-1,2]\]
\[f'(t)= \sqrt{4-t^2} + \frac{ t^2 }{ \sqrt{4-t^2 }}\]

Answer 2

Is my derivative right?

Answer 3

the derivative of 4-t^2 is (-2t)

Answer 4

oops i meant to put negative

Answer 5

nice then yes right

Answer 6

\[\frac{ 4 }{ \sqrt{4-t^2 }}\]

Answer 7

what's that ??

Answer 8

i'm trying to simplify to solve for t

Answer 9

how did you get 4 at the top ?

Answer 10

lcm

Answer 11

ohh wait...

Answer 12

\[f'(t)= \sqrt{4-t^2} -\frac{ t^2 }{ \sqrt{4-t^2 }}\]
when you apply chain rule you should get (-2t)
so that's how it is -t^2/sqrt(4-t^2)

Answer 13

Ohh so the c#'s are +- sqrt(2) ?

Answer 14

now get the common denominator
to find the critical points set it equal to 0

Answer 15

\[\frac{ 4-2t^2 }{ \sqrt{4-t^2} }=0\]

Answer 16

the numerator will make the rational equal to zero so I just look at the numerator and solve for t

Answer 17

watch out there is an interval

Answer 18

yes

Answer 19

Yes, I need to plug them into the function and see which are the absolute max/min

Answer 20

the question is asking critical points right ??
sqrt{2} is the only critical point here because [-1,2]

Answer 21

ohh ok :)

Answer 22

thank you :)

Answer 23

yw :)