Calc Question
Let g be the function with the derivative
$$g'(x) = 2x + \frac{ 16 }{ x ^{2} }$$determine the interval(s) for which the function g is increasing and concave up.

I was getting that increasing is (-2 to infinity)
and concave up is (negative infinity to -16^(1/3) )
but they don't overlap so there is no interval were it is both increasing and concave up.
would that be right or did I mess up somewhere?

Question

Calc Question
Let g be the function with the derivative
$$g'(x) = 2x + \frac{ 16 }{ x ^{2} }$$determine the interval(s) for which the function g is increasing and concave up.

I was getting that increasing is (-2 to infinity)
and concave up is  (negative infinity to -16^(1/3) )
but they don't overlap so there is no interval were it is both increasing and concave up.  
would that be right or did I mess up somewhere?

primeralph · Answer

Show your work and reasoning.

amoodarya · Answer

attachment

jigglypuff314 · Answer

I was looking at this http://www.wolframalpha.com/input/?i=2x+%2B+16%2Fx%5E2 which would be the graph of the derivative and when g'(x) is positive then g(x) is increasing

amoodarya · Answer

let see what you type !
you type it 2x+26/x^2
i solve it with this condition 
now you can replace 26 by 16

jigglypuff314 · Answer

whops sorry, that was a typo in the question
2x + 16/x^2

amoodarya · Answer

can you go on ?

jigglypuff314 · Answer

sorry, could you check again with the that  g'(x) = 2x + 15/x^2  pwease? :3

amoodarya · Answer

g'(x) = 2x + 15/x^2     or    g'(x) = 2x + 16/x^2

jigglypuff314 · Answer

crap Dx

2x + 16/x^2

amoodarya · Answer

attachment

jigglypuff314 · Answer

ohhh okay, I see what I did wrong, thank you!!!

amoodarya · Answer

your welcome