Answer
(a) $f'(a)=3a^2-2$
(b) $y=-2x+4$.
$y=x+2$.
$y=10x-12$
(c) See graph.
Work Step by Step
(a) Use the formula given in this section, we have:
$$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=\lim_{h\to 0}\frac{(a+h)^3-2(a+h)+4-a^3+2a-4}{h}=\lim_{h\to 0}\frac{ 3a^2h+3ah^2+h^3-2h}{h}=\lim_{h\to 0}(3a^2+3ah+h^2-2)=3a^2-2$$
(b) At $x=0$, we have $f(0)=4$ and $f'(0)=-2$ which is the slope of the tangent line. Knowing the slope and a point $(0,4)$ on the line, we can get the equation as $y=-2x+4$.
At $x=1$, we have $f(1)=3$ and $f'(1)=1$ which is the slope of the tangent line. Knowing the slope and a point $(1,3)$ on the line, we can get the equation as $y=(x-1)+3$ or $y=x+2$.
At $x=2$, we have $f(2)=2^3-4+4=8$ and $f'(2)=3\times2^2-2=10$ which is the slope of the tangent line. Knowing the slope and a point $(2,8)$ on the line, we can get the equation as $y=10(x-2)+8$ or $y=10x-12$
(c) See graph.