If g$ (x) = \max |y^2-xy|$ (0≤y≤1) then minimum value of $g (x)$ is?

Question

If g$ (x) = \max |y^2-xy|$ where (0≤y≤1) then minimum value of $g (x)$ (for real x ) is ?

Any suggestions for this sum ? I couldn't solve

Answer 1

If $x\ge2$, then $g(x)=\max y(x-y)=x-1\ge1$. If $1\le x\le2$, then the max is at $y=x/2$ giving $\frac{x^2}{4}\ge\frac{1}{4}$. If $0\le x\le 1$, then the max over $y\in[0,x]$ is $\frac{x^2}{4}$ and the max over $y\in[x,1]$ is $1-x$. So $g(x)=\max(1-x,x^2/4)$ which has the min value $3-2\sqrt2=0.171573$ over $[0,1]$.

If $x$ is negative, then the values are bigger, because $xy$ is positive. So the answer is $3-2\sqrt2$.

Answer 2

Hint: the maximum and minimum of $y\mapsto y^2-xy$ in $[0,1]$ will be the critical point of the function $y = x/2$ (only acceptable if $\in(0,1)$) or the border points of the interval $y = 0,1$. The values of the function in this points are: $(x/2)^2-x^2/2 = \cdots$, 0, $1-x$.