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

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Answer 1 · 2017-01-03T16:42:53

${[\frac{d}{d x} 4 x^{2}]}_{x = 2}^{2} = 16$ $[d/(dx) 4x^2]_(x=2) = 16$

By definition:

$f'(2) = lim_(t->2) (f(t)-f(2))/(t-2) = lim_(t->2) (4t^2-16)/(t-2)=lim_(t->2) 4(t^2-4)/(t-2)=lim_(t->2)4 ((t-2)(t+2))/(t-2)=lim_(t->2)4 (t+2)=16$

Answer 2 · 2017-01-03T17:10:06

$f'(2)=16$

$f(x)=4x^2$
Find f'(x) then plug in 2 for x.
$f'(x)=8x$
$f'(2)=8(2)$
$f'(2)=16$

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