Question

show that the inflection points for y=sinx/x lie on the curve y^2(x^4 +4)=4

Answer 1

set f''(x) =0, and find the infection point, whn you plug the x-solutions into the f(x).
by f(x) i am referring to y=sinx/x.

then check if these points are on the other function, y^2(x^4+4)=4.

Answer 2

I mean the x-solutions that result from f''(x)=0.
call them a and b,
you would have points,
(a,f(a)) and (b,f(b)).

Answer 3

the just plug in these (a,f(a)) and (b,f(b)) into y^2(x^4+4)=4..

Answer 4

so the second derivative would be [(2-x^2)sin(x) - 2xcos(x)]/x^3
right?

Answer 5

just over x, not over x^3, I think.

Answer 6

I got to go, sorry.