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

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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.!