Asympotes, Data?Bump
posted by Alaina on .
If someone could help me on this, I'd appreciate it.
f(x)= 2x/(x^2+1)
Vertical: none
Horizontal: none
Minimum: x=1
Maximum: x=1
Inflection Points: it seems like there are one or two, but i don't know how to factor f''(x)= 4x^3 + x^2  12x + 1 :/

Wolfram confirms the following:
http://www.wolframalpha.com/input/?i=2x%2F%28x%5E2%2B1%29+
No VA
No HA
max =1
min = 1
to get 2nd derivative
Wolfram said:
4x(x^2  3)/(x^2 + 1)^3
http://www.wolframalpha.com/input/?i=2%28x%5E2+1%29%2F%28x%5E2%2B1%29%5E2
So check your work for the second derivative
for points of inflection:
4x(x^2  3)/(x^2 + 1)^3 = 0
4x(x^2  3) = 0
so x = 0 or x^2 = 3
x = 0 or x + √3 or x = √3
if x = 0 , y = 0 > (0,0)
if x = √3 , y = 2√3/5 > (2√3/5)
if x = √3 > (2√3/5)
this is confirmed by looking of the graph in the first Wolfram link 
Thank you so much! I see what I should have done now. :P