Answer
$$\eqalign{
& \left( {\text{a}} \right){\text{ }}f'\left( 0 \right) = 0,{\text{ }}f'\left( {\frac{1}{2}} \right) = \frac{1}{4},{\text{ }}f'\left( 1 \right) = 1,{\text{ }}f'\left( 2 \right) = 4,{\text{ }}f'\left( 3 \right) = 9 \cr
& \left( {\text{b}} \right)f'\left( { - \frac{1}{2}} \right) = \frac{1}{4},f'\left( { - 1} \right) = 1,f'\left( { - 2} \right) = 4,f'\left( { - 3} \right) = 9 \cr
& \left( {\text{c}} \right){\text{The graph }}f'\left( x \right){\text{ is shown below}} \cr
& \left( {\text{d}} \right)f'\left( x \right) = {x^2} \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = \frac{1}{3}{x^3} \cr
& \left( {\text{a}} \right){\text{ }} \cr
& {\text{*Estimate }}f'\left( 0 \right){\text{, from the graph we can see that the}} \cr
& {\text{slope at }}x = 0{\text{ is 0}}{\text{, then }}f'\left( 0 \right) = 0 \cr
& {\text{*Estimate }}f'\left( {\frac{1}{2}} \right){\text{, from the graph we can see that the}} \cr
& {\text{slope at }}x = \frac{1}{2}{\text{ is }}\frac{1}{4}{\text{, then }}f'\left( {\frac{1}{2}} \right) = \frac{1}{4} \cr
& {\text{*Estimate }}f'\left( 1 \right){\text{, from the graph we can see that the}} \cr
& {\text{slope at }}x = 1{\text{ is }}1,{\text{ then }}f'\left( 1 \right) = 1 \cr
& {\text{*Estimate }}f'\left( 2 \right){\text{, from the graph we can see that the}} \cr
& {\text{slope at }}x = 2{\text{ is }}4,{\text{ then }}f'\left( 2 \right) = 4 \cr
& {\text{*Estimate }}f'\left( 3 \right){\text{, from the graph we can see that the}} \cr
& {\text{slope at }}x = 3{\text{ is }}9,{\text{ then }}f'\left( 3 \right) = 9 \cr
& \cr
& \left( {\text{b}} \right){\text{The function }}f\left( x \right) = \frac{1}{3}{x^3}\,{\text{is odd}}{\text{, and its derivative is}} \cr
& f'\left( x \right) = {x^2},{\text{ an even function}}{\text{, then by symmetry the }} \cr
& {\text{derivatives are equal}}{\text{.}} \cr
& {\text{*}}f'\left( { - \frac{1}{2}} \right) = f'\left( {\frac{1}{2}} \right) = \frac{1}{4} \cr
& {\text{*}}f'\left( { - 1} \right) = f'\left( 1 \right) = 1 \cr
& {\text{*}}f'\left( { - 2} \right) = f'\left( 2 \right) = 4 \cr
& {\text{*}}f'\left( { - 3} \right) = f'\left( 3 \right) = 9 \cr
& \cr
& \left( {\text{c}} \right){\text{The possible graph }}f'\left( x \right){\text{ is shown below}} \cr
& \cr
& \left( {\text{d}} \right){\text{By the definition of the derivative}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( {x + \Delta x} \right) - f\left( x \right)}}{{\Delta x}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\frac{1}{3}{{\left( {x + \Delta x} \right)}^3} - \frac{1}{3}{x^3}}}{{\Delta x}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\frac{1}{3}\left( {{x^3} + 3{x^2}\left( {\Delta x} \right) + 3x{{\left( {\Delta x} \right)}^2} + {{\left( {\Delta x} \right)}^3}} \right) - \frac{1}{3}{x^3}}}{{\Delta x}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\frac{1}{3}{x^3} + {x^2}\left( {\Delta x} \right) + x{{\left( {\Delta x} \right)}^2} + \frac{1}{3}{{\left( {\Delta x} \right)}^3} - \frac{1}{3}{x^3}}}{{\Delta x}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{{x^2}\left( {\Delta x} \right) + x{{\left( {\Delta x} \right)}^2} + \frac{1}{3}{{\left( {\Delta x} \right)}^3}}}{{\Delta x}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \left( {{x^2} + x\left( {\Delta x} \right) + \frac{1}{3}{{\left( {\Delta x} \right)}^2}} \right) \cr
& {\text{Evaluate when }}\Delta x \to 0 \cr
& f'\left( x \right) = {x^2} \cr} $$
