## Calculus with Applications (10th Edition)

$$f'\left( x \right) = 6{x^2},\,\,\,\,f'\left( { - 2} \right) = 24,\,\,\,\,f'\left( 0 \right) = 0,{\text{ and }}f'\left( 3 \right) = 54$$
\eqalign{ & f\left( x \right) = 2{x^3} + 5 \cr & {\text{Using the definition of the derivative}}: \cr & \cr & {\text{Find }}f\left( {x + h} \right) \cr & f\left( {x + h} \right) = 2{\left( {x + h} \right)^3} + 5 \cr & f\left( {x + h} \right) = 2\left( {{x^3} + 3{x^2}h + 3x{h^2} + {h^3}} \right) + 5 \cr & f\left( {x + h} \right) = 2{x^3} + 6{x^2}h + 6x{h^2} + 2{h^3} + 5 \cr & \cr & {\text{Find }}f\left( {x + h} \right) - f\left( x \right) \cr & f\left( {x + h} \right) - f\left( x \right) = 2{x^3} + 6{x^2}h + 6x{h^2} + 2{h^3} + 5 - \left( {2{x^3} + 5} \right) \cr & f\left( {x + h} \right) - f\left( x \right) = 6{x^2}h + 6x{h^2} + 2{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{{6{x^2}h + 6x{h^2} + 2{h^3}}}{h} \cr & \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = 6{x^2} + 6xh + 2{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( {6{x^2} + 6xh + 2{h^2}} \right) \cr & {\text{Then}}{\text{,}} \cr & f'\left( x \right) = 6{x^2} + 6x\left( 0 \right) + 2{\left( 0 \right)^2} \cr & f'\left( x \right) = 6{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) = 6{\left( { - 2} \right)^2} = 24 \cr & f'\left( 0 \right) = 6{\left( 0 \right)^2} = 0 \cr & f'\left( 3 \right) = 6{\left( 3 \right)^2} = 54 \cr}