Question

can't remember the exact question, but I was given a number: 17^(1/4), and told to find the linear proximation.

Answer 1

well what number close to 17 do we know the 4th root of exactly?

Answer 2

16^1/4

Answer 3

=2

Answer 4

good, so what function will we be approximating?

Answer 5

thats the thing, there was no function! It was part b of a question on my exam, but part A was totally unrelated to it... D:

Answer 6

the idea of a linear approximation is that we create a function that we know a known value of to estimate a value close to it

Answer 7

in this case we are asked to find the 4th root of a number
we know the 4th root of a number close to it
hence out function should be...?

Answer 8

exactly, I asked my prof because I thought there was a typo, but she smiled and said I was on the right track..

Answer 9

I don't understand..

Answer 10

we are approximating 4th roots, does it not make sense to make use of the function\[f(x)=x^{1/4}\]???

Answer 11

so, find the linear app. of that equation with the point given?

Answer 12

yes

Answer 13

so what x value is our approximation based around?

Answer 14

would that be the number given in the question??

Answer 15

no, we need to base our estimate around an x value \(x_0\) for which we already know f(x)
what is the x value that we already said we know f(x) of ?

Answer 16

16

Answer 17

yes, so \(f(x)=x^{1/4}\) and \(x_0=16\) for our purposes
now we just need to fill in the pieces to get our linear approximation function...

Answer 18

for \(x\approx x_0\) we have the linear approximation\[f(x)\approx f(x_0)+f'(x_0)(x-x_0)\]

Answer 19

so we need all the pieces of the formula...
what is \(f(x_0)\) ?

Answer 20

2

Answer 21

good
what is \(f'(x)\) ?

Answer 22

1/4(x^(-3/4))

Answer 23

good, so what is \(f'(x_0)\) ?

Answer 24

1/32?

Answer 25

excellent, not plug into the formula\[f(x)\approx f(x_0)+f'(x_0)(x-x_0)\]what do you get?

Answer 26

now*

Answer 27

x/32 + 3/2?

Answer 28

yep :)
now we can estimate f(17), which comes out to be...?

Answer 29

O.O you mean just plug it into the equation?! And get a 2.03??

Answer 30

yep, that's what we're doing :)

Answer 31

O.O did we just answer the question?!

Answer 32

I mean you did.. and me ish

Answer 33

yes you did; you really did all the work yourself, I just guided you
notice the logic just so it doesn't seem magic...

Answer 34

:D awesome thanks!

Answer 35

which means I got that question wrong on my test D:

Answer 36

for \(x\approx x_0\)\[f(x)\approx f(x_0)+f'(x_0)(x-x_0)\]in our case we got\[f(x)\approx\frac x{32}+\frac32\]so\[f(17)\approx\frac {17}{32}+\frac32=\frac{65}{32}\] since \(f(x)=x^{1/4}\) this statement says\[17^{1/4}\approx\frac{65}{32}\]which is what we wanted to know

Answer 37

the linear approximation gives 2.03125 and my calculator gives 2.03054...
so they are pretty darn close

sorry to hear about your test, better luck next time!

Answer 38

D: thanks anyways!
btw (y) cowboy bebop!

Answer 39

yeah, it's a classic!
Spike and I have the same last name too :)

Answer 40

O.O! sorry, had to check your profile, thats so cool

Answer 41

:)