Question

Please Help me out...

Answer 1

\[\large \bf 2x^2+\frac{1}{x}>0\]
Find the intervals of x

Answer 2

add

Answer 3

oh wow this is going to suck

Answer 4

\[\large \bf \frac{2x^3+1}{x}>0\]

Answer 5

\[\frac{2x^3+1}{x}>0\] is a start

Answer 6

then,

Answer 7

then we know the sign changes at \(x=0\) for the denominator

Answer 8

for the numerator it changes at \(-\frac{1}{\sqrt[3]{2}}\)

Answer 9

how?

Answer 10

please show your work

Answer 11

so the answer is nice and ugly
but no matter

Answer 12

\[2x^3+1=0\\
2x^3=-1\\
x^3=-\frac{1}{2}\\
...\]

Answer 13

then,here i stuck

Answer 14

take the cube root and get \(x=-\frac{1}{\sqrt[3]{2}}\)

Answer 15

yeah!!! thank you

Answer 16

oh, you mean stuck as to finding the answer to the original question? we are not done

Answer 17

what we are not done mean?

Answer 18

you still have to solve where it is positive, your answer should be intervals (2 intervals)

Answer 19

yeah... i can do that...

Answer 20

thanks..

Answer 21

k you want me to check, or is it obvious?

Answer 22

am i right or not?
answer=\[\large \bf (-\infty,\frac{-1}{\sqrt[3]{2}})U(0,\infty)\]