Question

use differentials or linear approximation to find an approximated value for 1/∛(8.2)

Answer 1

@JoelSjogren

Answer 2

the cube root of a 8 is easy to find right?
so...
let \[f(x)=\frac{1}{\sqrt[3]{x}}\]

Then find f'(x)
Then find f'(8)
Then say f(x) is approximately
f'(8)(x-8)+f(8)

let me know when you get this far

Answer 3

A hint: before differentiating \(f\), rewrite it as \(f(x) = \cfrac{1}{x^{1/3}} = x^{-1/3}\).

Answer 4

Now that I think of it, maybe that's unnecessary. What do you think?

Answer 5

I think it is a groovy hint.

Answer 6

Now you can apply power rule in that form.
I prefer that over quotient rule.

Answer 7

Basically we are finding the tangent line to that curve at x=8
For values near 8 we should get some pretty close approximations
but as we move further from the point that we found the tangent line that particular line actually becomes a bad approximation to use

Answer 8

f'(x)= -1/3x^4/3

Answer 9

that x^4/3 is also in the denominator right?

Answer 10

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

Answer 11

yes and f'(8)= -1/48

Answer 12

great so far

Answer 13

f(x) is approximately (-1/48)(x-8)+(1/2)

Answer 14

and what do we want to find
we want to find 1/cuberoot(8.2)

but f(x)=1/cuberoot(x)

so if 1/cuberoot(x) is approx (-1/48)(x-8)+1/2
then what do we want to plug in for x?

Answer 15

f(x)=(-1/48)x+(1/12)

Answer 16

you'd plug in 8.2?

Answer 17

right
and in the function approximation you wrote i think that 1/12 should be a 1/2 :)

Answer 18

oh you distribute sorry

Answer 19

so f(x)=(-1/48)(8.2)+(1/12) correct?

Answer 20

is it 2/3 instead of 1/12 though?

Answer 21

youre right. i multiplied 1/6 and 1/2 instead of adding them

Answer 22

and f(x) isn't approx that but
f(8.2) is approx that
so f(8.2) approx= -1/48*(8.2)+2/3

Answer 23

and then the rest is algebra

Answer 24

the cool thing about this way is you can find 1/cuberoot(8.2) without using a calculator

Answer 25

or at least approximate it

Answer 26

okay so i got 119/240 but how would you find 1/cuberoot(8.2) without a calculator?

Answer 27

i meant approximately by using the above way

Answer 28

I think the 2/3 at the end should be a 1/2.

Answer 29

119/240 is approximately 1/cuberoot(8.2)

Answer 30

no she distributed @aum

Answer 31

she did have \[f(x) \approx \frac{-1}{48}(x-8)+\frac{1}{2}\]

Answer 32

\[f(x) \approx \frac{-1}{48}x +\frac{8}{48}+\frac{1}{2}\]

Answer 33

you have to distribute, dont you?

Answer 34

\(\Large f'(a) = \lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a} \)

When x is very close to 'a', we can ignore the limit and use the following approximation:

\(\Large f'(a) \approx \frac{f(x)-f(a)}{x-a} \)
\(\Large f(x) \approx f'(a)*(x-a) + f(a) \)
\(\Large f(8.2) \approx f'(8)*(8.2-8.0) + f(8) \)

Answer 35

\[f(x) \approx \frac{-1}{48}x+\frac{1}{6}+\frac{1}{2}=\frac{-1}{48}x+\frac{2}{12}+\frac{6}{12}=\frac{-1}{48}x+\frac{8}{12} \\ \]
Therefore
\[f(x) \approx \frac{-1}{48}x+\frac{2}{3}\]

Answer 36

tiff you did everything right you found an approximation to the value you were asked to approximate using the linearzation method

Answer 37

okay, thank you so much for the help! math definitely is NOT my strong suite

Answer 38

you can see your approximation is pretty close since 119/240 is .4958333333333333333333(repeating)
and the actual value using the calculator which is just approximation anyways is .495901
these two values are pretty close

Answer 39

you did pretty good
you actually did the power rule which a lot people struggle with
I don't think you are as bad as you think