Question

Find the minimum

Answer 1

\[\LARGE f(x) =1 , x=0\\ \LARGE f(x) = x^{\sqrt{x}},0<x \le2\]

Answer 2

For f(x) the open interval [0,2]

Answer 3

logarithmic differentiation

Answer 4

yea I did that and got

f(x) = y[rootx/x + lnx/2rootx]

Answer 5

simplify that \[f'(x) =f(x)(\frac{ 1 }{ \sqrt{x} })[1+\frac{ 1 }{ 2 }lnx] \]

Answer 6

you need to put f'(x) = 0
right ?
so you just need 1+ 1/2 ln x = 0
for what values of x is that true ?

Also, you will need to test the endpoints too

Answer 7

Yea I did that and I got \[x=e^{-2}\]

Answer 8

so I pug that in and now I don't know how to simplify

Answer 9

\[f(e^{-2}) = (e^{-2})^{\sqrt{e^{-2}}}\]

Answer 10

\[=(e^{-2})^{e^{-2}\frac{1}{2}}\]

Answer 11

Is it \[e^{\frac{-2}{e}}\]

Answer 12

yea, so the exponent is -2 (e^-1) = -2/e

and e^{-2/e} = 0.4791 approx

Answer 13

alright wait did you guess that or use a calculator?

Answer 14

calculator

Answer 15

oh I though you knew how to estimate what e^-2 is..dang that would be a useful knowledge ...haha. thanks