$(-,-1)$

$(0,13)$ because $f^{\text{'}\text{'}}(x)$ is negative ant

Find the interval on which $f(x)=x\sqrt[\phantom{2}]{x+3}$ is increasing and concave up$.$

$f^{\text{'}}(x)=(x)(\frac{1}{2\sqrt[\phantom{2}]{x}+3})+(1)(\sqrt[\phantom{2}]{x+3})$

$\frac{x}{2\sqrt[\phantom{2}]{x}+3}+\frac{\sqrt[\phantom{2}]{x+3}}{1}\frac{2(x+3)+x}{2\sqrt[\phantom{2}]{x+3}}\frac{3x+6}{2\sqrt[\phantom{2}]{x+3}}=0,x=-2$

$f^{\text{'}\text{'}}(x)=$