# Asympotes, Data?-Bump

posted by .

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 :/

• Asympotes, Data?-Bump -

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

• Asympotes, Data?-Bump -

Thank you so much! I see what I should have done now. :P

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