optimization

[Arushi :))]

in case the interval is not given.. how to find the global maxima n minima over the entire domain????
f= (1+2/x)(x+6)^1/2
in this case why is -6 a local minima

[kangkan]

Hi arushi..the domain of f(x) is x>= -6 and x not equal to zero.

d/dx (fx)=(1-2/x^2)(x+6)^0.5 +(1+2/x)*0.5*(x+6)^0.5

We can clearly see that d/dx of f(x) vanishes at x=-2 and x=-6 among others. Hence -6 is a stationary point.
Now f(-6)=0. Also d/dx of F(x) >0 in the neighbourhood of x=-6..hence f(x) is increasing. Hence F(x) is local minimum....i hope this helps....i am not sure how to answer about global maximum and minimum, but lim f(x) as x->infinity is infinity,hence i dont think a global maximum exists

[Arushi :))]

yaar.. at -6 d/dx is infinity..that measns derivative doesnt exist..chek it again

[Arushi :))]

hen you find fx.. at x=6, -4.. f'x is zero
at -6 its infinity..
6 n -4 n stationery points.. but m nt getting how is -6 a local minima

[kangkan]

yes..sorry ..i miscalculated...it is infinity at x=-6...also,x is a boundary point..if you calculate f'(x) in the neighbourhood of x(5.99,5.98 etc),you will see that its positive..so when we come right from x=-6,the function is increasing....since,f(x) is undefined at x<-6, we can say that x=-6 is a local minimum.. i graphed it out to double check this time :)

[Arushi :))]

got it/.......thanks:)

[kangkan]

Welcome :)