Chapter 3 - The Derivative - 3.4 Definition of the Derivative - 3.4 Exercises - Page 176: 20

$$f'\left( x \right) = 12{x^2},\,\,\,\,f'\left( { - 2} \right) = 48,\,\,\,\,f'\left( 0 \right) = 0,{\text{ and }}f'\left( 3 \right) = 108$$

$$\eqalign{ & f\left( x \right) = 4{x^3} - 3 \cr & {\text{Using the definition of the derivative}}: \cr & \cr & {\text{Find }}f\left( {x + h} \right) \cr & f\left( {x + h} \right) = 4{\left( {x + h} \right)^3} - 3 \cr & f\left( {x + h} \right) = 4\left( {{x^3} + 3{x^2}h + 3x{h^2} + {h^3}} \right) - 3 \cr & f\left( {x + h} \right) = 4{x^3} + 12{x^2}h + 12x{h^2} + 4{h^3} - 3 \cr & \cr & {\text{Find }}f\left( {x + h} \right) - f\left( x \right) \cr & f\left( {x + h} \right) - f\left( x \right) = 4{x^3} + 12{x^2}h + 12x{h^2} + 4{h^3} - 3 - \left( {4{x^3} - 3} \right) \cr & f\left( {x + h} \right) - f\left( x \right) = 4{x^3} + 12{x^2}h + 12x{h^2} + 4{h^3} - 3 - 4{x^3} + 3 \cr & f\left( {x + h} \right) - f\left( x \right) = 12{x^2}h + 12x{h^2} + 4{h^3} \cr & \cr & {\text{Divide by }}h{\text{ to get }}\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \cr & \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \frac{{12{x^2}h + 12x{h^2} + 4{h^3}}}{h} \cr & \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = 12{x^2} + 12xh + 4{h^2} \cr & \cr & {\text{Let }}h \to 0 \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \left( {12{x^2} + 12xh + 4{h^2}} \right) \cr & {\text{Then}}{\text{,}} \cr & f'\left( x \right) = 12{x^2} + 12x\left( 0 \right) + 4{\left( 0 \right)^2} \cr & f'\left( x \right) = 12{x^2} \cr & \cr & {\text{Find }}f'\left( { - 2} \right),f'\left( 0 \right){\text{ and }}f'\left( 3 \right) \cr & f'\left( { - 2} \right) = 12{\left( { - 2} \right)^2} = 48 \cr & f'\left( 0 \right) = 12{\left( 0 \right)^2} = 0 \cr & f'\left( 3 \right) = 12{\left( 3 \right)^2} = 108 \cr} $$