$f\text{'}\text{'}\left(x\right)=3x+12=0$ gives you $x=-4$. Inflection point $(-4,-26)$.

$y\text{'}\text{'}\left(x\right)=8-6x^2=0$ gives you $x=\pm \sqrt{\frac{4}{3}}$. Inflection points $(\mathrm{-}\sqrt{\frac{4}{3}},\frac{40}{9})$ and $(\sqrt{\frac{4}{3}},\frac{40}{9})$.

$[-15,15]\times [-1500,1500]$

$f\left(x\right)=g\left(x\right)$ gives you $x=0\vee x=\pm \sqrt{\left(148\right)}$.
Solution: $\mathrm{-}\sqrt{148}<x<0\vee 0<x<\sqrt{148}$.

$g\text{'}\text{'}\left(x\right)=3x^2-72=0$ gives you $x=\pm \sqrt{24}$, so $(\pm \sqrt{24},-720)$.

$g\text{'}\left(\sqrt{24}\right)=-48\sqrt{24}$ and $g\text{'}\left(\mathrm{-}\sqrt{24}\right)=48\sqrt{24}$.
The intersect of the two inflection point tangents lies on the $y$ -axis, therefore $(0,432)$.

$TC\text{'}\text{'}=3q-8=0$ gives you $q=\frac{8}{3}$.
That is about $167$ kg.

$TP=q(11-q)-(\mathrm{0,5}{q}^3-4q^2+11q+4)=\mathrm{-0,5}{q}^3+3q^2-4$ and $TP\text{'}=\mathrm{-1,5}{q}^2+6q$.
$TP\text{'}=0$ when $q=0\vee q=4$.
The maximum profit is seen at a production volume of $400$ kg.

$x=-2$ and $x=3$

$x=\frac{1}{2}$ because the first derivative has a minimum at that value.

Negative, because $f\text{'}\left(\frac{1}{2}\right)<0$.

$f\text{'}\left(x\right)=4x^3+2ax=0$ gives you $x=0\vee x=\pm \sqrt{\mathrm{-}\frac{1}{2}a}$ and because $a>0$ this root cannot be a real number. Therefore only $x=0$ is a zero (x-intersect) of the derivative. Since $f\text{'}$ moves from negative to positive for $x=0$ then $f\left(0\right)$ must be a minimum.

$f\text{'}\text{'}\left(x\right)=12x^2+2a=0$ gives you $x=\pm \sqrt{\mathrm{-}\frac{1}{6}a}$ and because $a>0$ this root cannot be a real number. Therefore $f\text{'}\text{'}$ does not have any zeros (x-intersects) and thus no inflection points.