Answer
$$f'\left( x \right) = - 2,\,\,\,\,f'\left( { - 2} \right) = - 2,\,\,\,\,f'\left( 0 \right) = - 2,{\text{ and }}f'\left( 3 \right) = - 2$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = - 2x + 5 \cr
& {\text{Using the definition of the derivative}}: \cr
& {\text{Find }}f\left( {x + h} \right) \cr
& f\left( {x + h} \right) = - 2\left( {x + h} \right) + 5 \cr
& f\left( {x + h} \right) = - 2x - 2h + 5 \cr
& {\text{Find and simplify }}f\left( {x + h} \right) - f\left( x \right) \cr
& f\left( {x + h} \right) - f\left( x \right) = - 2x - 2h + 5 - \left( { - 2x + 5} \right) \cr
& f\left( {x + h} \right) - f\left( x \right) = - 2x - 2h + 5 + 2x - 5 \cr
& f\left( {x + h} \right) - f\left( x \right) = - 2h \cr
& {\text{Divide by }}h{\text{ to get }}\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \cr
& \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \frac{{ - 2h}}{h} \cr
& \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = - 2 \cr
& {\text{Let }}h \to 0 \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \left( { - 2} \right) \cr
& {\text{Then}}{\text{,}} \cr
& f'\left( x \right) = - 2 \cr
& \cr
& {\text{Find }}f'\left( { - 2} \right),f'\left( 0 \right){\text{ and }}f'\left( 3 \right) \cr
& f'\left( { - 2} \right) = - 2 \cr
& f'\left( 0 \right) = - 2 \cr
& f'\left( 3 \right) = - 2 \cr} $$
