Using only definition of derivative find $f'(a)$ when $f(x) = 2x^2 -5x + 1$

Question

Using only definition of derivative find $f'(a)$ when $f(x) = 2x^2 -5x + 1$

$$f'(a) = \lim_{x\to a} \frac{f(x) - f(a)}{x-a} = \lim_{x\to a} = \frac{2x^2 -5x + 1 - 2a^2 + 5a - 1}{x - a} = \lim_{x\to a} \frac{2x^2 - 2a^2 - 5x + 5a}{x-a}$$

What now?

Answer 1

$$f(x)=2x^2-5x+1$$ According to the definition of limit $$f'(a)=\lim_{x\to 0}\dfrac{f(a+h)-f(a)}{h}$$ $$\implies f'(a)=\lim_{x\to 0}\dfrac{\big(2(a+h)^2-5(a+h)+1\big)-(2a^2-5a+1)}{h}$$ $$\implies f'(a)=\lim_{x\to 0}\dfrac{2a^2+4ah+2h^2-5a-5h+1-2a^2+5a-1}{h}$$ $$\implies f'(a)=\lim_{x\to 0}\dfrac{2h^2+4ah-5h}{h}$$ $$\implies f'(a)=\lim_{x\to 0}\dfrac{h(2h+4a-5)}{h}$$ $$\implies f'(a)=\lim_{x\to 0}~~~2h+4a-5$$ $$So,~~~ f'(a)=4a-5$$

Answer 2

Note that: $2x^2-2a^2-5x+5a=2(x+a)(x-a)-5(x-a)$. Your next step is to find ways to remove your $x-a$ in your denominator. It is just a simple factorization. $$\frac{(x-a)[2(x+a)-5]}{x-a}=2(x+a)-5$$ Then, you can sub $x$ to be $a$ to solve it

Answer 3

Let $h=x-a$, and express the top line in terms of $a$ and $h$. Then it's easy $-$ no factorisation is required.