Question

Let f(x) be mx-1 + 1/x. Then the smallest value of the constant m such that f(x)>=0 for every x>0.

Answer 1

may be or maybe not....

Answer 2

try solving the equation f(x)=0 for x

Answer 3

constraint is there for x so cant

Answer 4

ignore the constraint for now

Answer 5

m x-1+1/x=0. multiply both sides by x

Answer 6

nope that will not work

Answer 7

once you solve for x in terms of m, all you need to find is a value of m for which the x values are almost imaginary. this occurs when the discriminant is zero.

Answer 8

it gives an open interval at that value, 'that value' is the answer...

Answer 9

so any idea bout calculus?

Answer 10

x={(1+sqrt(1-4m))/2m,(1-sqrt(1-4m))/2m}. for the discriminant to be zero, m must equal 1/4.

Answer 11

its in an open interval so cant say 1/4

Answer 12

you mean the interval for the x values?

Answer 13

what u are saying is x>1/4
so it will not include 1/4

Answer 14

for m>1/4, f(x)>0. thus, for m=1/4, f(x)>=0.

Answer 15

nope....cant say like that

Answer 16

ok sure. using calculus, set the first derivative equal to zero

Answer 17

to find the minimum point of the function

Answer 18

you get x=1/sqrt(m). now plug this value back into the original equation, and set f(1/sqrt(m)) equal to zero.

Answer 19

d/dx (m x - 1 + 1/x)=0
m-1/x^2=0
1/x^2=m
x=1/sqrt(m)

Answer 20

f(1/sqrt(m))=m/sqrt(m)-1+sqrt(m)=0.
m-sqrt(m)+m=0
2m-sqrt(m)=0
sqrt(m)=2m
m=1/4

Answer 21

thanks buddy...

Answer 22

:)