Question

Determine the intervals where the graph is concave up and concave down. y = x + 1/x

Answer 1

@electrokid can u help?

Answer 2

your second derivative is?

Answer 3

I got 2/x^3

Answer 4

and when that is equal to zero, OR is undefined, we get a value of x=?

Answer 5

no solutions exist

Answer 6

there is 1 solution that defines the parameters.

we know that 2 will never be equal to zero, so equating to 0 is out

we know the when the denominator of this is equal to 0 that we get an undefined. which is ONE of the parameters we need to satisfy

Answer 7

ok

Answer 8

so, we dont have an inflection at x=0, but we do have a vertical asymptote since the original setup is not defined for x=0 either.

now when you plug in a - value you get what?
and a + value gets you what?

Answer 9

plug in an a - value into x + 1/x ?

Answer 10

of course not, that function tells us nothing about concavity at a glance does it?

Answer 11

no

Answer 12

since the second derivative relates to concavity, lets try to use it instead

Answer 13

okay so plug in a - into 2/x^3

Answer 14

correct

Answer 15

ok so 2/-2^3 I get -1/4 and 2/2^3 = 1/4

Answer 16

great, so for any - value on into -inf gets us a - concavity

and any + value onward into +inf gets us a + concavity

and at the point x=0 it is simply undefined so we exclude x=0

Answer 17

ok

Answer 18

how would you create a set interval notation for such a property?

Answer 19

well so -1/4, infinity it will concave down and 1/4, infinity it will concave up right?

Answer 20

correct, and dont forget the split at 0

cave UP: (0,inf)
cave DOWN: (-inf,0)

Answer 21

okay

Answer 22

so that it right?

Answer 23

thats*

Answer 24

well, that is what we came to so yes i would have to say that is the correct results

Answer 25

ok