Answer
a) $ f′(x)=6x^2-1$
b) $5,5$
Work Step by Step
a) Now, $ f′(x)=\lim_\limits{ h\to 0} \dfrac{f(x+h)-f(x)}{h}$
or, $=\lim_\limits{ h\to 0} \dfrac{[2(x+h)^3-(x+h)]-(2x^3-x)}{h}$
or, $=\lim_\limits{ h\to 0} \dfrac{6x^2h+6xh^2+2h^3-h}{h}$
or, $=6x^2+6x(0)+2(0)^2-1$
or, $ f′(x)=6x^2-1$
b) $ f′(-1)=6 (-1)^2-1=5$
and $ f′(1)=6 (1)^2-1=5$
