N(t)=0.81t-1.14square root t + 1.53. Find absolute extrema on [0,6.
derivative -0.19 - 0.57/t^.5
the only critical value is 0 (right? check please)
I found N(6) = 3.60
The answers page also has N(.5) I don't see where the came from.

Question

N(t)=0.81t-1.14square root t + 1.53. Find absolute extrema on [0,6. 
derivative -0.19 - 0.57/t^.5 
the only critical value is 0 (right? check please) 
I found N(6) = 3.60
The answers page also has N(.5) I don't see where the came from.

anonymous · Answer

$$N(t)=.81t-1.14\sqrt{t+1.53}$$ i hate decimals

anonymous · Answer

or did i get the function wrong?

anonymous · Answer

the square root is only over the t

anonymous · Answer

$$N(t)=.81t-1.14\sqrt{t}+1.53$$

anonymous · Answer

that's it

anonymous · Answer

derivative is 
$$.81-\frac{.57}{\sqrt{t}}$$

anonymous · Answer

let me check my math again

anonymous · Answer

in any case i think this is increasing over your interval, and i don't see any .5 in the problem

anonymous · Answer

oh no i am wrong
there is a critical point in there

anonymous · Answer

it is at $t=\frac{361}{729}$ which is not really .5 but close enough i guess

anonymous · Answer

since you are working with annoying decimals anyway, might as well round

anonymous · Answer

how'd you get .81 instead of -.19? for your derivative

anonymous · Answer

you wrote $.81t$ and the derivative of $.81t$ is $.81$

anonymous · Answer

i was going to ask you where the $-.19$ came from

anonymous · Answer

it was on the ans page

anonymous · Answer

thanx

anonymous · Answer

yw