Answer
a) $ f′(x)=6x+12$
b) $0, 18$
Work Step by Step
a) Now, $ f′(x)=\lim_\limits{ h\to 0} \dfrac{f(x+h)-f(x)}{h}$
or, $=\lim_\limits{ h\to 0} \dfrac{[3(x+h)^2+12(x+h)-1]-(3x^2+12x-1)}{h}$
or, $=\lim_\limits{ h\to 0} \dfrac{6xh+3h^2+12h}{h}$
or, $=3(2x+4)$
or, $ f′(x)=6x+12$
b) $ f′(-2)=6 \times (-2) +12=0$
and $ f′(1)=6 \times (1) +12=18$
