$f(x)=-(x-2k)^2(x-4k)-k\quad$ where $k\gt 0\quad$

Question

$f(x)=-(x-2k)^2(x-4k)-k\quad$ where $k\gt 0\quad$

(i) Show that the local max occurs at $\left(\frac{10k}3,\frac{32k^3}{27}-k\right)\quad$.
(ii) If the local max point is to occur on the x-axis, find the exact value of k.

Hi, I proved that the first part is correct. However, I don't know how to do the second question. Can anyone help me where should I start.

Answer 1

This is a cubic with negative $x^3$ term, so if it has a local minimum and a local maximum, then the local maximum occurs at the larger value of $x$. Expanding, we find the cubic is $-x^3+8kx^2-20k^2x+16k^3-k$. Differentiating we get $-3x^2+16kx-20k^2=-(3x-10k)(x-2k)$.

We are given that $k>0$, so the local maximum occurs at $x=\frac{10k}{3}$. The value of the polynomial at that point is $-k+\frac{32k^3}{27}$. That is zero for $k=0,\pm\sqrt{\frac{27}{32}}$. But $k>0$, so that gives us the solution $k=\sqrt{\frac{27}{32}}$.