Absolute mininum value of y=x+1/x on the inerval x>0 occurs when x=1/x? - what is the inerval between blood donnations
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Monday, February 8, 2010
What Is The Inerval Between Blood Donnations Absolute Mininum Value Of Y=x+1/x On The Inerval X>0 Occurs When X=1/x?
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Its function is
ReplyDeletey = x + 1 / x
To find the miniumum to set the derivative to zero:
0 = dy / dx = 1 -1 / x ^ 2
1 / x ^ 2 = 1
x ^ 2 = 1
x = 1 (-1 out of range)
And its minimum value is (1) = 1 +1 / 1 = 2
Answer 2
Agebra another solution.
ReplyDeletefor all, the number B
this is real
(| AB |) ^ 2 must be equal to 0 and more
then
(| AB |) ^ 2 = 0
A ^ 2 + b ^ 2-2AB = 0
A ^ 2 + b ^ 2 = 2AB
or
2AB = \\ \\ \\ \\ \\ \\ \\ \\ u0026lt, A ^ 2 + b ^ 2
Question x 1 / x
type = A x B = 1 / x
Then 2ab = \\ \\ \\ \\ \\ \\ \\ \\ u0026lt, A ^ 2 + b ^ 2
2 (x) (1 / x) = \\ \\ \\ \\ \\ \\ \\ \\ u0026lt, x ^ 2 + (1 / x) ^ 2
2 = \\ \\ \\ \\ \\ \\ \\ \\ u0026lt, x ^ 2 + (1 / x) ^ 2
the minimum of x ^ 2 + (1 / x) ^ 2 = 2
if we know the minimal value, we can solve for x
x ^ 2 + (1 / x) ^ 2 = 2
x = 1 or -1, but x> 0
then x = 1
Return to the question
y = x 1 / x
to x = 1
So y = x 1 / x = 1 +1 = 2 is minimal
... ANSWER