Question

Can someone help me with this problem please

Answer 1

can you help me

Answer 2

what do you think, give me your idea first, it's not a piece of cake

Answer 3

um I think that we have to put the e^k where ever we see an x first

Answer 4

how about the "global minimum value of fx"?

Answer 5

yeah that is where I get confused with these problems

Answer 6

is it not the vertex of parabola? since it's is upward, so the minimum value is the vertex of it .

Answer 7

ok

Answer 8

so then 2

Answer 9

ok

Answer 10

f'(x)=x+2xlnx

Answer 11

=0 when?

Answer 12

?

Answer 13

is it right with the derivative?

Answer 14

yes

Answer 15

sorry, you are right,

Answer 16

so then what?

Answer 17

and that derivative =0 to get the x value of the vertex,

Answer 18

=0 when?

Answer 19

so then add 0 where there is an x

Answer 20

no, it =0 when x(1+2lnx)=0 iff x =0 or 2lnx =-1 (reject)

Answer 21

oh ok

Answer 22

sorry, not that, not reject, reject x =0

Answer 23

accept the value of 2lnx = -1 ---> lnx = -1/2 ---> x = e^-1/2

Answer 24

combine with yours x = e^k ----> k = 1/2

Answer 25

got it?

Answer 26

yes

Answer 27

ohoh -1/2

Answer 28

why -.5

Answer 29

sorry, so tired , really got?

Answer 30

yeap

Answer 31

so then -.5 is the final answer

Answer 32

yeap

Answer 33

I think so. if not, I hand off,

Answer 34

thanks