MHB Maximizing $y=|4x^3+ax^2+bx+c|$ in $[-1,1]$

[anemone]

Let $a,\,b$ and $c$ be real numbers and let $K$ be the maximum of the function $y=|4x^3+ax^2+bx+c|$ in the interval $[-1,1]$. Show that $K\ge 1$. For which $a,\,b$ and $c$ is the equality occurs?

 

[Opalg]

Spoiler

Let $p(x) = -ax^2 - bx - c$, so $p(x)$ is a quadratic (or lower degree) polynomial, and we want to find the smallest possible value of $K = \max\{|4x^3 - p(x)|:-1\leqslant x\leqslant 1\}.$

Notice that if $|4x^3 - p(x)| \leqslant K$ for $-1\leqslant x\leqslant 1$, then $|4(-x)^3 - p(-x)| = |4x^3 + p(-x)| \leqslant K$ for $-1\leqslant x\leqslant 1$. Therefore $|4x^3 - q(x)| \leqslant K$ for $-1\leqslant x\leqslant 1$, where $q(x) = \frac12(p(x) - p(-x))$. But $q(x)$ is an odd function. So to find the function that minimises $K$ we need only look at odd functions, and the only odd polynomials of degree at most $2$ are multiples of $x$.

Therefore in the polynomial $p(x) = -ax^2 - bx - c$ we should take $a=c=0$, and it is easy to see that the optimum value of $b$ is $-3$. Then (as in the diagram below) the graph of $y= 4x^3$ lies between the lines $y = 3x\pm1$ in the interval $[-1,1]$, and $1 = \max\{|4x^3 - 3x|:-1\leqslant x\leqslant 1\}$ is the smallest possible value for $K$.

[TIKZ][xscale=4,yscale=5]
\draw [step=0.25cm, help lines] (-1.1,-1.1) grid (1.1,1.1) ;
\draw (-1.1,0) -- (1.1,0) ;
\draw (0,-1.1) -- (0,1.1) ;
\draw[very thick, domain=-1.1:1.1] plot (\x,{\x^3});
\draw [thick] (-1.1, -0.825) -- (1.1,0.825) ;
\draw [dashed,thin] (-1.1, -1.075) -- (1.1,0.575) ;
\draw [dashed,thin] (-1.1, -0.575) -- (1.1,1.075) ;
\foreach \x in {-1,0.5,1} \draw (\x,-0.07) node [fill=black!8] {$\x$} ;
\foreach \y in {-4,-3,...,4} \draw (-0.13,\y/4) node [fill=black!8] {$\y$} ;[/TIKZ]