Question

find absolute minimum at (x,y)

f(x) = 2 (square root x) (x − 1); x ≥ 0

Answer 1

2*x^(1/2) * (x-1) = 2*x^(3/2) - 2*x^(1/2)
Take derivative: (6/2)(x^(1/2)) - x^(-1/2)
Set equal to zero to find critical point: (6/2)(x^(1/2)) - x^(-1/2) = 0
(6/2)(x^(1/2)) = x^(-1/2)
\[\frac{ 6\sqrt{x} }{ 2 }=\frac{ 1 }{ \sqrt{x} }\]
Multiply both sides by sqrt(x):
\[\frac{ 6x }{ 2 } = 1\]
\[6x = 2\]
\[x = \frac{ 1 }{ 3 }\]
Plug x into original equation to get:
\[\frac{ -4\sqrt{1/3} }{ 3 }\]

Answer 2

@carson889 i just plugged that into my assignment and it stated it was incorrect. did you write down the problem correctly?

Answer 3

\[2\sqrt{x}(x-1) ; x \ge0\]

Answer 4

nvm i got it. thanks for your help

Answer 5

Your welcome. And glad it worked out.