How do you sketch the graph `y=1/(1+x^4)` using the first and second derivatives?

To analyze the behavior of the function we start from considering the domain of definition and see that `y(x)` is defined and continuous in all of `RR` as the denominator never vanishes.

We can also see that `y(x)` is positive in its domain and it it even as

`y(-x) = y(x)`

so that its graph is symmetrical with respect to the `y`-axis.

At the limits of the domain `y(x)` is infinitesimal as:

`lim_(x->+-oo) y(x) = 0`

so the function is going to have `y=0` as asymptote on both sides.

Now we calculate the first and second derivatives:

`y'(x) = - (4x^3)/(1+x^4)^2`

`y''(x) = frac ( -12x^2(1+x^4)^2 +32x^6(1+x^4)) ((1+x^4)^4) = frac ( -12x^2(1+x^4) +32x^6) ((1+x^4)^3) = frac (4x^2( -3x^4 +8x^4-3) )((1+x^2)^3) = frac (4x^2(5x^4-3)) ((1+x^2)^3)`

We can see that the only critical point where `y'(x) = 0` is for `x=0` and that:

`x<0 => y'(x) >0`
`x>0 => y'(x) <0`

so that the function is monotone increasing for `x in (-oo,0)`, reaches a maximum for `x=0` and then is monotone decreasing for `x in (0,+oo)`.

The second derivative has three zeros in `x=0` and `x=+-root(4)(3/5)`. We know the first is a maximum, so `y(x)` has two inflection points in `x=+-root(4)(3/5)`. and:

`abs(x) > root(4)(3/5) => y''(x) > 0` and y(x) is concave up.
`abs(x) < root(4)(3/5) => y''(x) < 0` and y(x) is concave down.

graph{1/(1+x^4) [-10, 10, -5, 5]}