Is g(x) = 1/(x^2+1) Concave Down on Which Intervals?

[vmtyler]

Homework Statement

For what intervals is g(x) = 1/(x^2+1) concave down?

Homework Equations

Quotient Rule

The Attempt at a Solution

OK, so I have found the first and second derivative, but I am confused as to how I should use them to find where the function is concave down. Do I set the second derivative equal to zero? Use the quadratic formula? I know the answer from the back of the book, but my attemps to get the same answer have failed. Please point me in the right direction! Thank you :)

 

[gabbagabbahey]

Setting the second derivative equal to zero will give you the inflection points, but you also need to determine whether the second derivative is (+) or (-) on each side of these inflection points...what are you getting for your solution?

 

[vmtyler]

Ok, so for the 1st derivative I have:

-2x/(x^2+1)^2

Second:

-6x^2-2/ (x^2+1)^2

Do I need to factor the denom? Also, you stated that I need to determine the values on either side of the inflection points, I should do that with my calculator, right? So, if I have this correct, determining the value (+ or -) of the function on both sides of the inflection points will tell if the original funtion is concave up or down?

Thank you!

 

[gabbagabbahey]

Close, I get the numerator as being +6x^2-2... and you don't need to factor the denominator; the expression will only be zero when the numerator is zero...what does that give you for you inflection points?

 

[vmtyler]

WOW! You have been so much help!

I got


X= sqrt of (1/3)

So the function is concave down for all X values between -< X < + of sqrt (1/3)

I have been going crazy trying to figure this is out! Thank you for your time :)

 

[gabbagabbahey]

No problem