Question

i need help please. the question is posted as a link that i took a picture and uploaded. any help would be appreciated.
http://prntscr.com/4wvey3

Answer 1

ok,. what are your thoughts on part a?

Answer 2

we have an outside function (...)^(1/3); being controlled by an inside function 4x+1

what process do we know of that deals with function controlling functions?

Answer 3

umm ... composite functions ?

Answer 4

yeah, thatll be topical ... i was looking more for something called a chain rule

Answer 5

can you define the chain rule for me?

Answer 6

taking the derivative of the outside and then the derivative of the inside ?

Answer 7

yes, and multiplying the results

[f(g)]' = f'(g) * g'

Answer 8

in this case: f = (...)^1/3 , g = 4x+1

what is f' and g'?

Answer 9

f' = 1/3(9/10)^-(2/3)
g' = 4

Answer 10

let me chk

g' is fine

f' is a power rule

1/3 (...)^(-2/3)

youve got the form right .... lets test the value

f'(g(0)) * g'(0) = 1/3 (4(0)+1)^(-2/3) * 4

f'(g(0)) * g'(0) = 4/3

Answer 11

this gives us the slope of the tangent line at the point (0,f(g(0))) which i spose ironically is a y intercept

Answer 12

well i am confused. what did you do once we found f' and g'

Answer 13

i used the rule that i defined prior to it

[f(g(x))]' = f'(g) * g'

but i was bad and used f for a generality while the problem used it as a specific function

Answer 14

at any rate ..

f'(g) = 1/3 g^(-2/3)
g' = 4

putting it all together

f'(g)*g' = 4/3 g^(-2/3)


now we defined the g function as 4x+1, g=1 when x=0

Answer 15

ok so how do we use the linearization to estimate this then ?

Answer 16

recall that the derivative of a function defines the slope of a tangent line at a given point

with a point and a slope we can define the equation of a line

Answer 17

better yet, when x=0 we have a y intercept and can use the more common slope intercept form of a line

Answer 18

yes i understand. but how does that help use to calculate the linear approximation

Answer 19

by creating the line equation of course

L(x) = mx + b

or if we want to use the terms of the problem using calculus

L(x) = f'(x) (x-0) + f(x)

Answer 20

well, in this case since its at the point when x=0

L(x) = f'(0) (x-0) + f(0)

Answer 21

ok. the question also says " using this linearization" what does that mean ? does that mean we have to use something from part (a) of the question ?

Answer 22

it means use L(x) to find an approximation of f(x) when x=9/10

Answer 23

L(x) = f'(0) (x-0) + f(0)

let x = .9

L(.9) = f'(0) (.9-0) + f(0)

Answer 24

ok i understand till L(.9) = f'(0)(.9-0)+f(0)

now how do i calculate this value

Answer 25

i mean what function am i putting 0.9 into ?

Answer 26

well, i already calculated f'(0) and f(0) is pretty straight forward since they gave you f(x)

Answer 27

L(x) is the function that linearizes f(x) so we use L(x) to approximate f(x) with

Answer 28

L(x) = (4/3)x + 1 is that right ?

Answer 29

that is the part (a) answer i got ^

Answer 30

thats correct for part b yes

Answer 31

part b yes sorry

Answer 32

so i just put 0.9 into this function to get the answer for c ?

Answer 33

yep

Answer 34

so the answer to the whoooole part c question is 1 ?

Answer 35

4/3 (9/10) + 1 is not going to be 1

12/10 + 1

1.2 + 1

Answer 36

ohh ok that clarifies things a lot. thanks again amistre :)

Answer 37

good lcuk