Question

show that the curve has 3 points of inflection and they all lie on 1 straight line:\[y=\frac{1+x}{1+x^{2}}\]

Answer 1

so to find my inflection point i let y''=0?

spent the whole evening and i got \[x=1,x=\pm \sqrt{3}-4\]is it correct?

my 3rd derivative for the equation is \[y'''=\frac{-6(x^{4}+4x^{3}-6x^{2}-6x+1)}{(1+x^{2})^{4}}\]
so now i substitute my x into my y''' to make sure it doesnt goes to 0?

Answer 2

no it's wrong. just find the first derivative and equate it's numerator to zero. you will get
2x^3 -x+1=0
which have three roots and those are the points of inflection where dy/dx is zero

Answer 3

ou i thought we need y''' not equal to zero to prove its a point of inflection

Answer 4

you already got three points,
the get y value from y = (1+x)/(1+x^2), find the slope, if the slopes are equal then they line in the same line.

Answer 5

yea but im not sure whether my points are correct. they are not in the same line, ie slopes are different.and also when im trying to prove x=-rt(3) -4 to be an inflection point, y''' is -0.00075 which is approximately 0, so it shouldnt be an inflection point? lol. not sure of my own point... T_T

Answer 6

You want to find the zeros of the second derivative. Not the first or third.

Answer 7

yea i used the zeros of 2nd derivative. x=1,x=±√3−4
but when i substitute -√3−4into y''' to check for inflection point, it approximates to 0.

Answer 8

Those aren't the zeros I'm getting for the 2nd derivative

Answer 9

o.o, so the 3 points for the x coordinate is correct?

Answer 10

oh wait == x=±√3−2, not 4

Answer 11

That looks better. Use those 3 x-coordinates, and find the y-coordinates of those points on the original function.

Answer 12

oh kay. thx. lol i hate myself == wasted so many papers on this

Answer 13

You should see the notebooks I've filled with math scribblings.

Answer 14

lol. i feel so dumb.. finals in 2more days cant even finish the revision.