Review question

# When does $y=kx^3-(k+1)x^2+(2-k)x-k$ have a minimum at $x =1$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8441

## Solution

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.