Question

Given the function `f(x)= 1/12x^4 + 1/6x^3-3x^2-2x+1` how do you find any points of inflection and determine where the curve is concave up or down?

Answer 1

If `f` is twice differentiable on an open set A, the inflections are the points where `f^{''}` changes sing. This is a polynomial so it's obvious that it's smooth (i.e. infinitively differentiable) on `RR`, so here, the inflection points are the points where `f^{''}=0` and exist `n>1` such that `f^((2n+1))!=0`.

So `f^{''}=x^2 +x -6` and its roots are
`{-1 +- sqrt(1+24)}/2 = {2, -3}`

Now we notice that the third derivative: `2x+1` has zero only in 1, so it's not zero in the candidate inflection points, and this tells us that -3 is a falling point and 2 is a rising point

(this does make sense since we know how the graphic of a quartic polynomial behave)
Now, if the function is twice differentiable, the concavity is determined by the sign of its second derivative:
`f^{''}>0 in U:= {x<-3 or x>2}`
`f^{''}<0 in D :={-3 < x <2}`

So in U the curve is concave up, and in D the curve is concave down