When does $y=kx^3-(k+1)x^2+(2-k)x-k$ have a minimum at $x =1$?

When does $y=kx^3-(k+1)x^2+(2-k)x-k$ have a minimum at $x =1$?

Question

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

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

Suggestion

How can we find the turning points on a graph?

How can we find out what kind of turning points they are?

Solution

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.

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/kbTxRppv/width/463/height/469/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

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