Question

How do you determine where the graph of the given function is increasing, decreasing, concave up, and concave down for `h(x) = (x^2) / (x^2+1)`?

Answer 1

First we need to evaluate the domain and the first and second derivatives:
`D_h=RR`
`h'(x)=(2x)/(x^2+1)^2`
`h''(x)=(-2(3x^2-1))/(x^2+1)^3=(-6(x+sqrt(3)/3)(x-sqrt(3)/3))/(x^2+1)^3`

Now, `h` is increasing when `h'>0` and decreasing when `h'<0`. Notice that `forall_(x in RR)\ x^2+1>0` so as long and we're interested only in the sing of the derivative we can ommit the denominator.
`2x>0 iff x>0` - function `h` increases when `x in (0;+oo)`
`2x<0 iff x<0` - function `h` decreases when `x in (-oo;0)`

Function `h` is concave up when `h''>0` and concave down when `h''<0`.
concave up: `-6(x+sqrt(3)/3)(x-sqrt(3)/3)>0 iff x in (-sqrt(3)/3;sqrt(3)/3)`
concave down: `-6(x+sqrt(3)/3)(x-sqrt(3)/3)<0 iff`
`iff x in (-oo;-sqrt(3)/3) cup (sqrt(3)/3;+oo)`