Question

I have an answer can anyoe tell me what they got.
Find the absolute maximum and minimum for y=x^2+(2/x) on 0.3 < or equal to x < or equal to 4

Answer 1

what did you get?

Answer 2

for max I got-4 min I got 8

Answer 3

Is that correct

Answer 4

now, check the domain :)

Answer 5

\[0.3\le x\le4\]

Answer 6

sorry but I a a little confused

Answer 7

hold it..
thats not it!!!

Answer 8

\[y=x^2+2x^{-1}\\
y'=2x-2x^{-2}=0\\\implies2x-{2\over x^2}=0\implies x^3-1=0\implies\boxed{x=1}
\]
there is only one critical value.. (note that x=0 is not in the domain of the function)

Answer 9

ok

Answer 10

so the the max is -1

Answer 11

how did you get the "-1"??

Answer 12

sorry I meant1

Answer 13

to check if it is a maxima or a minima, we need the second derivative..
so, find the second derivative

Answer 14

ok it would be 2x+4x^-1=0

Answer 15

\[y'=2x-2x^{-2}\\y''=?\]

Answer 16

it would be \[-\frac{ 12 }{ x^4}\]

Answer 17

\[y''=2+4x^{-3}\]