Question

How do you determine whether the function `f(x)= 4/(x^2+1)` is concave up or concave down and its intervals?

Answer 1

`f(x)` is concave up for `|x| > 1/sqrt5` and concave down for `|x| < 1/sqrt5`

Explanation:

If a function is differentiable twice, we know that it is concave up if `f''(x) >0` and concave down viceversa.

So let us calculate:

`f(x) = 4/(x^2+1) = 4(x^2+1)^(-2)`

`f'(x) = -16x(x^2+1)^(-3)`

`f''(x) = 96x^2(x^2+1)^(-4)-16(x^2+1)^(-3) = frac(96x^2-16(x^2+1)) ((x^2+1)^4)= frac(96x^2-16x^2-16) ((x^2+1)^4)=16frac (5x^2-1) ((x^2+1)^4)`

Evidently the denominator of `f''(x)` is always positive, so the sign of `f''(x)` is the sign of the numerator and we see that:

`f''(x) < 0` for `|x|< 1/sqrt(5)`

So `f(x)` is concave up in the intervals `(-oo,-1/sqrt(5))` and `(1/sqrt(5),+oo)` and concave down for `x` in the interval `(-1/sqrt(5),1/sqrt(5))`.

graph{4/(x^2+1) [-10, 10, -5, 5]}