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?

1 Answer
Apr 21, 2015

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