To define the concavity we must find the derivatives of the function:
[tex]\begin{gathered} f(x)=x\sqrt{x^2+4} \\ f^{\prime}(x)=(x^{\prime})(\sqrt{x^2+4})+(x)(\sqrt{x^2+4})^{\prime} \\ f^{\prime}(x)=1(\sqrt{x^2+4})+x(\frac{x}{\sqrt{x^2+4}}) \end{gathered}[/tex][tex]\begin{gathered} f^{\prime}(x)=(\sqrt{x^2+4})+(\frac{x^2}{\sqrt{x^2+4}}) \\ f^{\prime}(x)=\frac{x^{2}}{x^{2}+4}+\frac{\sqrt{x^2+4}\sqrt{x^2+4}}{\sqrt{x^2+4}} \\ f^{\prime}(x)=\frac{x^2+(\sqrt{x^2+4})^2}{\sqrt{x^2+4}} \\ f^{\prime}(x)=\frac{x^2+x^2+4}{\sqrt{x^2+4}} \\ f^{\prime}(x)=\frac{2x^2+4}{\sqrt{x^2+4}} \end{gathered}[/tex]Now, we will continue finding the second derivative:
[tex]\begin{gathered} f^{\prime}(x)=\frac{2x^2+4}{\sqrt{x^2+4}} \\ f^{\prime\prime}(x)=\frac{{}\lbrack(2x^2+4)^{\prime}\sqrt{x^2+4}\rbrack-\lbrack(\sqrt{x^2+4})^{\prime}2x^2+4\rbrack}{(\sqrt{x^2+4})^2} \\ f^{\prime\prime}(x)=\frac{\lbrack4x(\sqrt{x^2+4})\rbrack-\lbrack\frac{x}{\sqrt{x^2+4}}\times2x^2+4\rbrack}{x^2+4} \end{gathered}[/tex][tex]\begin{gathered} f^{\prime\prime}(x)=\frac{4x(\sqrt{x^2+4})-\frac{x(2x^2+4)}{\sqrt{x^2+4}}}{x^2+4} \\ f^{\prime}^{\prime}(x)=\frac{\frac{4x(\sqrt{x^2+4})(\sqrt{x^2+4})}{\sqrt{x^2+4}}-\frac{x(2x^2+4)}{\sqrt{x^2+4}}}{x^2+4} \\ f^{\prime\prime}(x)=\frac{\frac{4x(x^2+4)-x(2x^2+4)}{\sqrt{x^2+4}}}{x^2+4} \end{gathered}[/tex][tex]\begin{gathered} f^{\prime\prime}(x)=\frac{\frac{4x^3+16x-2x^3-4x)}{\sqrt{x^2+4}}}{x^2+4} \\ f^{\operatorname{\prime}\operatorname{\prime}}(x)=\frac{\frac{2x^3+12x}{\sqrt{x^2+4}}}{x^2+4} \\ f^{\prime}^{\prime}(x)=\frac{2x^3+12x}{x^2+4(\sqrt{x^2+4})} \end{gathered}[/tex]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:
[tex]f^{\prime\prime}(x)=\frac{2x^3+12x}{(x^2+4)(x^2+4)}=0[/tex]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:
[tex]-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.can be:
[tex]x=-1\text{ and x=1}[/tex][tex]\begin{gathered} for\text{ }x=-1 \\ f´´(x)=-\frac{14\sqrt[\placeholder{⬚}]{5}}{25} \\ for\text{ }x=1 \\ f´´(x)=\frac{14\sqrt{5}}{25} \end{gathered}[/tex]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
