Question

Find a linear approximation of the function f(x)=1(1+2x)4 at a=0.

I got: 1+x
this is not right
what did i do wrong?

Answer 1

what exactly is your function?
\[ 1(1+2x)^4 \]?
what is there a 1 out front?
what does a=0 mean?

Answer 2

\[1/(1+2x)^4\]

Answer 3

that was the problem I didn't write it right

Answer 4

f(x)= f(a) + f'(a)(x-a)

OK,
what did you get for the derivative

Answer 5

i got
1+x

Answer 6

\[ f(x)= \frac{1}{(1+2x)^4} = (1+2x)^{-4} \]

this is (sort of) like \(u^{n}\) you should get \( n u^{n-1} du \)

Answer 7

let u= 1+2x

Answer 8

the derivative is not 1+x

Answer 9

im not sure how to solve this problem

Answer 10

can you take the derivative of
\[ u^{-4} \]

Answer 11

is it -8(1+2x)^-5

Answer 12

If you do not know how to take the derivative of, for example
\[ d (x^2) = 2x dx \]
this might help...
http://www.khanacademy.org/math/calculus/differential-calculus/v/calculus--derivatives-1

Answer 13

yes,
\[ \frac{d f(x)}{dx}= \frac{-8}{(1+2x)^5} \]

Answer 14

now you use
f(x)= f(a) + f'(a)(x-a) with a=0

what is f(0) ?

Answer 15

is the answer
1+-8x?

Answer 16

what is f(0)?
f(0)= 1/(1+2*0)^4 = ?

Answer 17

i mean
1-8x

Answer 18

are you asking about the final answer?

Answer 19

yes

Answer 20

you need f(0), f'(0) and (x-0)

f(0)= 1
f'(0)= -8
x-0 is x

plug into
f(x) = f(0) + f'(0)(x-0)

you get
f(x) = 1 + -8*x or
f(x)= 1-8x

Answer 21

sweet thanks for the help