Question

use the linear aproximation L(x) of the function f(x)= 5th root of x to the vaule 5th root of 31. find an x interval on which aproximation is with .1 of the actual vaule. use L(x) = f(a)+f'(x)(x-a)

Answer 1

well, we know the 5th root of 32, so use the approximation near the point x=32

Answer 2

that is, a=32

Answer 3

that should be
L(x) = f(a)+f'(a)(x-a)
btw

Answer 4

this practice exam has x in there. and why do we use 32 for a?

Answer 5

linear approximations are done around a known value x=a. we can find the 5th root of 32 easily, so let's use that point since it's near 31

Answer 6

f'(32) = 1/5 *32?

Answer 7

no, what is f'(x) ?

Answer 8

the derivitive of f(x)

Answer 9

yeah I know, I mean what is the derivative of f(x) in this case?

Answer 10

\[f \prime(x)=\sqrt[5]{32}\]

Answer 11

no, that is f(x)

Answer 12

actually that's f(32)
first off, what is f(x) ?
then what is the derivative f'(x) ?
finally what is f(32) and f'(32) ?

Answer 13

f(x) is 5th root of x
f'(x) wouls be 1/5x^(4/5)

Answer 14

f'(x)=1/5x^(-4/5)

Answer 15

so what are f(32) and f'(32) ?

Answer 16

if you move it to the denominator it becomes positive tho right?

Answer 17

yes, I didn't know it was supposed to be in the denom

Answer 18

5th root of 32 is 2
f'(32) is 1/80

Answer 19

yes, so what is L(x) ?

Answer 20

L(x)=2+(1/80)(x-32)

Answer 21

good, so what is L(31) ?

Answer 22

159/80

Answer 23

or 1.9875

Answer 24

and that makes sense, being a little less than 2
so we're done

Answer 25

awesome that made it much more clear to me, thank you

Answer 26

welcome!