The cubic \[y=kx^3-(k+1)x^2+(2-k)x-k\]

has a turning point, that is a minimum, when \(x = 1\) precisely for

\[\text{(a)} \quad k>0, \quad \text{(b)} \quad 0<k<1, \quad \text{(c)} \quad k > \dfrac{1}{2}, \quad \text{(d)} \quad k < 3, \quad \text{(e) all values of } x.\]

What happens if we differentiate \(y\)? We find \(\dfrac{dy}{dx} = 3kx^2-2(k+1)x+2-k\).

If we substitute \(x = 1\) into this, we find

\[\dfrac{dy}{dx} = 3k -2k -2 +2 -k = 0.\]

Thus we have a stationary point at \(x = 1\) whatever the value of \(k\). But when is this a minimum?

Differentiating again, we find that \(\dfrac{d^2y}{dx^2} = 6kx -2k -2\). At \(x = 1\), this gives \(\dfrac{d^2y}{dx^2}= 4k-2\).

The turning point will be a minimum if \(\dfrac{d^2y}{dx^2} > 0 \iff k > \dfrac{1}{2}\).

Thus the answer is (c).

Since our curve is a cubic, \(\dfrac{d^2y}{dx^2} = 0\) would always mean a point of inflection so is excluded from the inequality.

For higher order curves this isn’t always the case – consider \(y=(x-1)^4\) at the point \(x=1\), for instance.

This applet illustrates the shape of our cubic as \(k\) varies.