Let $f(x)=\frac{1}{9}{x}^3.$ Find a value $c$ between $x=0$ and $x=3$ so that the average tate of change of $f(x)$ from $x=0$ to $x=3$ is equal to the instantaneous rate of change of $f(x)$ at $x=c.$ You thay use the finct that $f^{\text{'}}(x)=\frac{1}{3}{x}^2.$

Possibilities:

$($a$)2$

$($b$)0$

$($c$)9$

$($d$)\sqrt[\phantom{2}]{2}$

$($e$)\sqrt[\phantom{2}]{3}$