Curver sketching

[nameVoid]

in trying to sketch the curve of y=(2-x^2)/(1+x^4)



y'=(-2x-8x^3+2x^5)/(1+x^4)^2

y''=[ (1+x^4)^2(-2-24x^2+10x^4)-(-2x-8x^3+2x^5)8x^3(1+x^4) ] / (1+x^4)^4

=[ -2(1+x^4)(1+12x^2-5x^4)+16x^4(1+4x^2-x^4) ] /(1+x^4)^3

dead end

 

[lanedance]

to get started try plotting 2-x^2, (1+x^4) and 1/(1+x^4), this should give you a feel for the general shape of the function

noticing that the denominator is always > 0 and both the numerator & denominator are symmetric should also help

 

[nameVoid]

i need to calculate the exact points of inflections

 

[lanedance]

hmmm... how about considering a substitution u = x^2 then

this should hopefully lead to a simpler calculation of the zeroes of the derivatives w.r.t. x

then if ' denotes dervative w.r.t. x, using the chain rule

$$y' = \frac{dy}{dx} = \frac{d}{dx} y(u(x)) = \frac{d y(u)}{du} \frac{du(x)}{dx} = \frac{dy}{du} u'$$

and so on for the next one, where you'll need the product rule as well