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

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Answer 1 · 2017-01-03T16:42:53

#[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#