On what interval(s) is the graph of f(x)=x/(x^2+1) concave down?

Answer 1

f(x) = x / (x^2 + 1)

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

f "(x) = 2x(x^2 - 3) / (x^2 + 1)^3

Inflection points occur when f "(x) = 0.

Let f "(x) = 2x(x^2 - 3) / (x^2 + 1)^3 = 0
Therefore, 2x(x^2 - 3) = 0,
so, x = 0 or x = -sqrt(3) or x = +sqrt(3)
These are the points where the function changes concavity.

Concave down occurs when f "(x) is negative.

Which amounts to : 2x(x^2 - 3) < 0

This occurs twice, at x < -sqrt(3) and 0 < x < sqrt(3).

Answer 2

f'[x]= (x^2+1)-x(2x)/ (x^2+1)^2
f'[x]= -x^2+1/(x^2+1)^2

f''[x]= -2x(x^2+1)^2 - (-x^2+1)(2(x^2+1)*2x) / (x^2+1)^4
f''[x]= -2x(x^2+1)^2 - (-x^2+1)(4x(x^2+1) / (x^2+1)^4

To find inflection points and concavity, set f''[x]=0, so
-2x(x^2+1)^2 - (-x^2+1)(4x(x^2+1) / (x^2+1)^4=0
-2x(x^2+1)^2 - (-x^2+1)(4x(x^2+1)=0
Solving, you'll get inflection point at x=0

Solving for concave up/down ,you get
-up-|-down-
......0

Therefore, concave down at x > 0

Answer 3

[0,∞)

Answer 4

xy=(-1,.5),(-1,1/2).