To define the concavity we must find the derivatives of the function:


Now, we will continue finding the second derivative:


![\begin{gathered} f^(\prime\prime)(x)=((4x^3+16x-2x^3-4x))/(√(x^2+4)))/(x^2+4) \\ f^{\operatorname{\prime}\operatorname{\prime}}(x)=((2x^3+12x)/(√(x^2+4)))/(x^2+4) \\ f^(\prime)^(\prime)(x)=(2x^3+12x)/(x^2+4(√(x^2+4))) \end{gathered}]()
Now we will find the roots of the first and second derivatives so that we will have the critical points:
Due to in the first derivative there is no solution for y=0 in the reals, we have no first-order critical points.
For the second derivate:

As we can see, the second derivative becomes zero when x=0, being the only critical point of second order.
We proceed to replace x=0 in the function f(x), and obtain as a result f(0)=0, being (0,0) the coordinate of the critical point.
Taking into account the given interval and the value of the second order critical point obtained, we have two new intervals:
![-4\leq\text{ x }<0\text{ and 0 }we investigate the concavity of the function in these intervals, taking test values within each interval. We are only interested in the sign and apply the criterion of the second derivative.<p></p><p>can be:</p>[tex]x=-1\text{ and x=1}]()
![\begin{gathered} for\text{ }x=-1 \\ f´´(x)=-\frac{14\sqrt[\placeholder{⬚}]{5}}{25} \\ for\text{ }x=1 \\ f´´(x)=(14√(5))/(25) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/od9uj35kcklicg2fkn6nl2j2mtyfibrwqh.png)
By the criterion of the second derivative, we can deduce that the function is concave downward in the interval [-4,0) because it gave negative and is concave upward in (0,6) because it gave positive.
So the answers are:
F(x) is concave down on the interval x=-4 to x=0
F(x) is concave up on the interval x=0 to x=6
The inflection point for this function is at x= 0
The minimum occurs at x=-4
The maximum occurs at x=6