How do you find f'(2) using the limit definition given f(x)=4x2?

By definition:f′(2)=limt→2f(t)−f(2)t−2=limt→24t2−16t−2=limt→24t2−4t−2=limt→24(t−2)(t+2)t−2=limt→24(t+2)=16Answer:[ddx4x2]x=2=16

Explanation:f(x)=4x2Find f'(x) then plug in 2 for x.f'(x)=8xf'(2)=8(2)f'(2)=16

By definition:f′(2)=limt→2f(t)−f(2)t−2=limt→24t2−16t−2=limt→24t2−4t−2=limt→24(t−2)(t+2)t−2=limt→24(t+2)=16Answer:[ddx4x2]x=2=16

Explanation:f(x)=4x2Find f'(x) then plug in 2 for x.f'(x)=8xf'(2)=8(2)f'(2)=16

Expert answers to "How do you find f'(2) using the limit..."

Saul Cochran

2022-04-17

How do you find f'(2) using the limit definition given

f (x) = 4 x^{2}

uvredio0of6

Beginner2022-04-18Added 17 answers

By definition:

f^{'} (2) = lim_{t \to 2} \frac{f (t) - f (2)}{t - 2} = lim_{t \to 2} \frac{4 t^{2} - 16}{t - 2} = lim_{t \to 2} 4 \frac{t^{2} - 4}{t - 2} =

lim_{t \to 2} 4 \frac{(t - 2) (t + 2)}{t - 2} = lim_{t \to 2} 4 (t + 2) = 16

Answer:

{[\frac{d}{d x} 4 x^{2}]}_{x = 2} = 16

prangijahnot

Beginner2022-04-19Added 17 answers

Explanation:

f (x) = 4 x^{2}

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

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?