Assume k is a constant. the function
g(x)= x4- 6k(x2) + 6k2 + x
has at least one point of inflection
a- if and only if k>0
b- if and only if k >= 0
My answer:
g'(x) = 4x3 - 12kx + 6k2
g''(x) = 12x2 - 12k
= 12(x2 - k)
k has to be greater than 0 to make g''(x) have any roots. But can't it also equal 0 making an inflection point at 0. I would have said B but the answer is A. Could someone please help me. Thanks!!!
It must be (a) because if k=0, then
g'' does not change signs, so there is no inflection. It remains concave up.