How do you find f'(x) using the limit definition given #f(x)=4+x-2x^2#?

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Answer 1 · 2016-07-22T14:24:03

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

Let us remember the defn. of #f'(x)=lim_(trarrx)[(f(t)-f(x))/(t-x)]#

Now, #f(x)=4+x-2x^2rArr f(t)=4+t-2t^2#

#rArr f(t)-f(x)=4+t-2t^2-4-x+2x^2#

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

#=(t-x){(1-2(t+x)}#

#rArr[(f(t)-f(x))/(t-x)]={(1-2(t+x)}, if, t!=x#.

We know that as #trarrx, t!=x#. Therefore,

#f'(x)=lim_(trarrx)[(f(t)-f(x))/(t-x)]#

#=lim_(trarrx){(1-2(t+x)}=1-2(x+x)=1-4x#.