Question

Find the linearization L(x) of the function f(x)=x^3/4 at x=256.

Answer 1

i got (3/16)x-16. the computer said it was wrong

Answer 2

@phi

Answer 3

thats what i did

Answer 4

sorry 64+(3/16)(x-256)

Answer 5

maybe they want you to simplify that

Answer 6

i did i got (3/16)x-16

Answer 7

how did you get -16 ?

Answer 8

when you expand 64+(3/16)(x-256 you get 64+(3/16)x-48 right?

Answer 9

yes, and 64-48 = +16

Answer 10

never mind its 16

Answer 11

oh my i am really failing today

Answer 12

The good news is you know how to do the hard part.

Answer 13

haha true. thanks

Answer 14

fixing the easy stuff is not that hard. Just slow down, and double check your arithmetic

Answer 15

i have another question. its telling me to use linear approximation to approximate (8.07)^(2/3). i got 4.023. it says i am wrong though

Answer 16

f(x) = x^(⅔)
f'(x) = (⅔) x ^(-⅓)
expand around x0= 8
f(8)= 4
f'(8) = ⅔ * ½= ⅓
f(x)= f(x0) + f'(x0)* (x-x0)
f(x) = 4 + ⅓ * (x-8)

f(8.07)= 4+ ⅓(.07)= 8.02333...

my calculator says 8^(⅔) is 4.023299437

Answer 17

yeah none of that works

Answer 18

did you try more digits ? 4.02333

Answer 19

yup

Answer 20

See if they have any instructions on how many digits they want. If you can't find that, make a note and ask your teacher, because the answer is 4.023 (3 repeating)

Answer 21

this is the question "Use a linear approximation to estimate the number (8.07)^2/3."

Answer 22

ok thanks

Answer 23

there is an outside chance there is a typo, and they meant 8.075
(because 0.075 is evenly divisible by 3) you would get 4.025

Answer 24

nope its not that. do you think you could help me out on one last problem

Answer 25

and just to be sure, ⅔ is the exponent, and it is not (8.07^2 ) * 1/3

Answer 26

its 2/3

Answer 27

Find the linearization L(x) of the function g(x)=xf(x2) at x=2 given the following information.
f(2)=−1 f′(2)=8 f(4)=4 f′(4)=−2

Answer 28

is that
\[ g(x) = x f(x^2) \]
? if so, use an ^ to show an exponent

Answer 29

yeah its that. my bad

Answer 30

find the usual stuff: g(2) and g'(2)
what do you get ?

Answer 31

g(2)=8 but i dont know how to find g'(x). would g'(x)=f(x^2)+2x^2f(x^2)?

Answer 32

product rule:
g(x) = x f(x^2)
g' = x f'(x^2) * 2x + f(x^2)
or
g'= f(x^2) + 2 x^2 f'(x^2)

yes, it looks the same as what you got.
now evaluate at x=2

Answer 33

it would be 36

Answer 34

how?

Answer 35

4+8(4)=4+32=36

Answer 36

g'= f(x^2) + 2 x^2 f'(x^2)
f(4) + 2* 2^2 * f'(4)

Answer 37

=36

Answer 38

f(2)=−1 f′(2)=8 f(4)=4 f′(4)=−2
isn't f'(4) = -2 ?

Answer 39

oh my. i cant get anything right. so it would be -12

Answer 40

FINALLY!!!!! i got it its -12x+32. thanks so much

Answer 41

Excellent!