Question

differentiation question:
draw the curve y=ln(x+1) for x values between 0 & 4 included. by drawing an appropriate straight line, obtain an approximate value for the positive square root of the equation x-2ln(x+1)=0.

Answer 1

i can't say i understand what this question wants. especially the part of about the "positive square root of
\[x-2\ln(x+1)\] which frankly makes no sense

Answer 2

however, if you are trying to solve
\[x-\ln(x+1)=0\] for x, then
\[x=0\] works by inspection

Answer 3

makes sense, but what about the second part?

Answer 4

that is the second part

Answer 5

isnt that just x-2y=0

Answer 6

here is the picture of
\[y=\ln(x+1)\] on the interval [0,4]
http://www.wolframalpha.com/input/?i=ln%28x%2B1%29+x%3D0+to+4

Answer 7

I used wolfram too :P
but what i dont get is the positive square root of the equation. how do i find that?

Answer 8

primary square root maybe?

Answer 9

what is this "square root" business?!
it is not a square

Answer 10

\[\pm sqrt{...}, \color{#dd0011}{\text{use the +}}\]

Answer 11

dunno the business part lol

Answer 12

\[x-2\ln(x+1)=0\] has two solutions, but in no sense are they "square roots" of anything

Answer 13

\[ \sqrt{x-2ln(x+1)};\ x-2ln(x+1)\ge0\]

Answer 14

and how this relates to the first question is beyond me. i thought it was going to be a visualization of newton's method, but it is not. the first expression is not the same as the second one

Answer 15

maybe it is supposed to read
"obtain an approximate value for the positive root of the equation x-2ln(x+1)=0. "

Answer 16

approximate: ln(x+1) with an appropriate line ... yada yada

Answer 17

maybe

Answer 18

yeah but that doesn't help solve
\[x-2\ln(x+1)=0\] as far as i can see

Answer 19

the equation of the tangent line to ln(x+1) at x gives as a close approx of ln(x+1)

Answer 20

maybe we need to use the approximation of ln(x+1) in its place

Answer 21

or maybe its just the average rate of change from 0 to 4 that we use

Answer 22

http://openstudy.com/?F225231636612OYKJOD=_#/updates/4eec41e9e4b0367162f56dc3
might be more info here

Answer 23

no, those are separate questions

Answer 24

y = .4024 x is the line drawn from 0 to 4

http://www.wolframalpha.com/input/?i=y+%3D+%28ln%285%29-ln%281%29%29x%2F%284%29and+ln%28x%2B1%29+x%3D0+to+4

Answer 25

if we use that in place of ln(x+1) we can get an approx of the equation instead of having to use ln(x+1)

Answer 26

you can get a better and better approx by using like the first 5 terms of its taylor series expansion

Answer 27

oh okay thanks guys!

Answer 28

thats the best I can make sense of it, good luck with it :)