Saul Cochran

2022-04-17

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

uvredio0of6

Beginner2022-04-18Added 17 answers

By definition:

${f}^{\prime}\left(2\right)=\underset{t\to 2}{lim}\frac{f\left(t\right)-f\left(2\right)}{t-2}=\underset{t\to 2}{lim}\frac{4{t}^{2}-16}{t-2}=\underset{t\to 2}{lim}4\frac{{t}^{2}-4}{t-2}=$

$\underset{t\to 2}{lim}4\frac{(t-2)(t+2)}{t-2}=\underset{t\to 2}{lim}4(t+2)=16$

Answer:

${\left[\frac{d}{dx}4{x}^{2}\right]}_{x=2}=16$

Answer:

prangijahnot

Beginner2022-04-19Added 17 answers

Explanation:

$f\left(x\right)=4{x}^{2}$

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

f'(x)=8x

f'(2)=8(2)

f'(2)=16

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

f'(x)=8x

f'(2)=8(2)

f'(2)=16

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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