Find the equation of the osculating circle at the local minimum of

Question

f(x)=2x^3+2x^2+(−5/6)x−8

Answer 1

find that x=1/6 and x= -5/6 make y'=0

at x=1/6, y''>0 so that point is local minimum (remember how y=x² works with y''=2 and has minimum)

x=1/6, y=1722/216

looks daunting, but look at curvature formula

K=y''/(1+y'²)^(3/2)

in this case simply K=y'' at x=1/6

K=6

so radius of circle is R=1/6

Must have radius perpendicular to curve at x=1/6 an y=1722/216

since slope is zero, means radius line is vertical and center is R up from y=1722/216

which is not so daunting

circle equation

(x-1/6)²+(y-[(-1722/216)+1/6])²=(1/6)²