How do you find f'(x) using the limit definition given $f (x) = 2 x^{2} - x$ $f(x)= 2x^2-x$ ?

Question

1461 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-09T04:22:34

$f'(x)=4x-1$ .

The Defn. $: f'(x)=lim_(trarrx)(f(t)-f(x))/(t-x)$ .

Given $f(x)=2x^2-x rArr f(t)=2t^2-t$ .

$:. f(t)-f(x)=2t^2-t-2x^2+x=ul(2t^2-2x^2)-ul(t+x)$

$=2(t^2-x^2)-(t-x)=2(t-x)(t+x)-(t-x)$

$=(t-x){2(t+x)-1}$ .

$:. (f(t)-f(x))/(t-x)={2(t+x)-1}, if tnex$ .

Therefore,

$f'(x)=lim_(trarrx)(f(t)-f(x))/(t-x)$

$=lim_(trarrx) {2(t+x)-1},...............[as, trarrx, tnex]$

$=2(x+x)-1$

$:. f'(x)=4x-1$ .

Enjoy Maths.!

How do you find f'(x) using the limit definition given f(x)= 2x^2-xf(x)=2x2−x f(x)= 2x^2-x?