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

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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,

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

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

How do you find f'(x) using the limit definition given f(x)=4+x-2x^2f(x)=4+x−2x2f(x)=4+x-2x^2?