## Calculus: Early Transcendentals 8th Edition

$$f'(a)=6a-4$$
$$f(x)=3x^2-4x+1$$ That means $$f(a)=3a^2-4a+1$$ According to definition, the derivative of $f(x)$ at number $a$ is $$f'(a)=\lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}$$ $$f'(a)=\lim\limits_{h\to0}\frac{[3(a+h)^2-4(a+h)+1]-[3a^2-4a+1]}{h}$$ $$f'(a)=\lim\limits_{h\to0}\frac{[(3a^2+3h^2+6ah)-4a-4h+1]-3a^2+4a-1}{h}$$ $$f'(a)=\lim\limits_{h\to0}\frac{3h^2+6ah-4h}{h}$$ $$f'(a)=\lim\limits_{h\to0}(3h+6a-4)$$ $$f'(a)=3\times0+6a-4$$ $$f'(a)=6a-4$$