Question

How do you do this problem? The question is attached as an image.

Answer 1

do you have notes on Linear Approximation?

Answer 2

yeah i know that L(x) = f(a) + f'(a)(x-a)

Answer 3

ok, that makes sense.
you need to find f(a) which means f(1) here
you also need the derivative of f(x)

Answer 4

f'(X) = 1/2 (8+X)^(-1/2)

Answer 5

yes, now evaluate that at x=1

Answer 6

i got .16666667

Answer 7

1/6

Answer 8

so far you know
\[ \frac{df}{dx} = \frac{1}{6} \]
that means
\[ \frac{\Delta f}{\Delta x} \approx \frac{1}{6} \]
and
\[ \Delta f \approx \frac{\Delta x}{6} \]

Answer 9

the derivative using "infinitesimals" is accurate
and approximate change small (but not infinitely small) \( \Delta f\)
is approximately the change in x divided by 6
plug in \( \Delta x = -1 \) and figure out the first answer.

Answer 10

ok so -1/6

Answer 11

now for the approximation at a=1
\[ L(x) = f(1) + \frac{1}{6} (x-1) \]
what did you get for f(1) ?

Answer 12

3

Answer 13

so L(x) = 3 + (x-1)/6

here , (x-1) represents the change in x , i.e. \( \Delta x\)
and they want \( \Delta x = -1 \)
in other words, x-1 = -1 or x=0
so they want you to find (approximately)
L(0)

notice when x is 0 in f(0), we are estimate \( \sqrt{8+0} = \sqrt{8} \)

Answer 14

okay i see. so the next part of the question is 8

Answer 15

yes.

Answer 16

though this problem is not written in the clearest way. I think it's very confusing.

Answer 17

wait so what would the next part of the question be. it's not -1/6 right?

Answer 18

and i know this is hard

Answer 19

not exactly. use the linear approximation formula, but replace (x-a) with -1

Answer 20

Here is a graph showing y= sqr(x), and the linear approximation (dotted blue line) that is the tangent to the red curve at x=9

Answer 21

L(x) = 3 + (x-1)/6
find
3 + -1/6

Answer 22

17/6

Answer 23

sorry how did you know to replace with -1 again?

Answer 24

they are making this very confusing. But they say \( \Delta x = -1\)

Answer 25

ohh ok

Answer 26

Here is the "big picture"
the derivative gives you a formula to find the slope of the tangent line at any point on the curve.

this problem is saying (in a very muddled way)
find the slope of the tangent at x=9 for the curve y= sqr(x)
the point on the curve at x=9 is (9, 3)
the slope is 1/6
the equation for the tangent is
y - 3 = 1/6(x-9)
using the point - slope formula
or
y = 3 + 1/6(x-9)

now find the *estimate* for sqrt(8) by finding the y value of the *line* (not the sqr curve)
y= 3 + 1/6(8-9)
y= 3 + 1/6(-1)
y= 3 - 1/6

Answer 27

you get y= 17/6= 2.833333....
compare that to the exact value of sqr(8)= 2.828427125..

Answer 28

so the The error in Linear Approximation is: .0049058753

Answer 29

how do you find the error in percentage terms?

Answer 30

relative error is
(estimate - truth)/truth

then multiply by 100 to change that fraction to a percent.

Answer 31

i got .1734606563 % but the computer says it's wrong

Answer 32

maybe they want you to round it ?
try
0.17
without the percent sign

Answer 33

no it's not taking it

Answer 34

Do they explain percentage error ? in your notes?

Answer 35

no the only percent error that i've ever known was the one you provided

Answer 36

it looks like a bug. try either 0.0017346 or 17.346

Answer 37

no neither of them work

Answer 38

ask your teacher about this one

Answer 39

ok that's fine

Answer 40

thank you so much. you explained it way clearer than me professor did :)

Answer 41

*my

Answer 42

you might try 0.4906 as a wild guess (of course this is totally wrong)

Answer 43

no it's not working

Answer 44

it's unlikely we can figure out what they want, if they don't want the correct answer.
0.173460667

Answer 45

nope :(

Answer 46

time to move on

Answer 47

lol yeah

Answer 48

thanks again!

Answer 49

yw