If $M (x)$ = max ($4-x$, ($\frac{\sqrt x^{3}}{\sqrt3^3}$) ) , $0< x \leq 4$ find minimum value of $M(x)$ in $(0 ,4]$.
what i did was first of all we need to find the x possible values for which $4-x$ is greater than the other given in the $M(x)$ function. Similarily for other . By solving two inqualities we would get the where x lies for each of them as a maximum . But after one finds those values and also the equality case , we would need to check at every point where function changes its form and also at end points . Is there a another approach to this without needing to consider case work and then checking for values at some critical points ?
One of those unusual Calculus problems that is best attacked without Calculus.
At $x=3,$ you have that $\displaystyle (4-x) = 1 = \left[\frac{\sqrt{x}}{\sqrt{3}}\right]^3.$
For $x < 3,$ the Max function will be controlled by
$(4-x) > 1.$
For $x > 3,$ the Max function will be controlled by
$\displaystyle \left[\frac{\sqrt{x}}{\sqrt{3}}\right]^3 > 1.$
Therefore, the minimum value of $(1)$, which can not be improved in the interval $0 \leq x < 4,$ is achieved at $x=3.$