inconsistent derivative.

[luckyday]

Hi guys,

I'm new to the forum and would love some help. I'm trying to find the relative max and min of x^2/(x^2 +1), as well as the inflection points.

y' came out to 2x/(x^2+1)^2. This leaves zero as the only critical value. When I pulled the second derivative, I got:

y''=(((x^2+1)^2)2 - 2x(2)(x^2+1)(2x))/((x^2+1)^2)^2

simplified

y''=(2-8x^2)/(x^2+1)^3

Unfortunately my book got y''= 2(1-3x^2)/(x^2+1)^3

This is my hold up. Could someone take a second derivative from x^2/(x^2+1) and tell me how you got it? Thanks so much! I'll try to help other folks if I can.

The way I worked it, I got x=0 as a relative min and plus and minus 1/2 as the inflection points but these are different from the answers in the book. I'm using the book 3000 + solved calculus problems. This is problem 15.10

 

[tkhunny]

It looks like you have it. The problem must be in your algebra.

 

[luckyday]

Hmm, I don't understand. I'm getting a different answer from the book. So unless the book's wrong...

What answer do you get for the second derivative?

Thanks for the reply

 

[galactus]

It would appear you have a small calculation error.

Quotient rule:

\(\displaystyle \frac{(x^{2}+1)(2)-2x(4x(x^{2}+1))}{(x^{2}+1)^{4}}\)

\(\displaystyle \frac{2(x^{2}+1)^{2}-8x^{2}(x^{2}+1)}{(x^{2}+1)^{4}}\)

Factor \(\displaystyle 2(x^{2}+1)\) from the numerator:

\(\displaystyle \frac{2(x^{2}+1)((x^{2}+1)-4x^{2})}{(x^{2}+1)^{4}}\)

\(\displaystyle \frac{2(1-3x^{2})}{(x^{2}+1)^{3}}\)

 

[luckyday]

Thank you, galactus. I appreciate the help.