Given $c > 0$ and $u > l > 0$, $$ \max_{x \in \Bbb R} \, \left( 1 - 2 c x^2 \right)^2 \quad \text{subject to} \quad l \leq x \leq u $$ Can the maximum value be found in terms of $l$ or $u$?
My try:
One can write the problem as follows: $$ \min_{ \begin{aligned} x&> 0\\ x- l &\geq 0\\ u -x &\geq 0 \end{aligned} } -(1-2cx^2)^2 $$ and define Lagrange multiplier $\lambda_1, \lambda_2, \lambda_3 \geq 0$ and consider 8 different situations for $\lambda$'s.
Question
Is there a hacky way of solving the above without considering constrained optimization?
Using the constraint $0<l≤x≤u$, we have:
$$1-2cu^2≤1-2cx^2≤1-2cl^2$$
This implies that,
$$ \begin{align}\max \left\{\left(1-2cx^2\right)^2\wedge 0<l≤x≤u\right\}=\max\left\{\left(1-2cu^2\right)^2,\;\left(1-2cl^2\right)^2\right\}\end{align} $$