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 ?

  • 1
    $\begingroup$ Have you tried sketching the functions? $\endgroup$
    – DatBoi
    Commented May 13, 2022 at 4:31
  • $\begingroup$ No thats why i didnt solve uaing the solution posted below @DatBoi $\endgroup$ Commented May 14, 2022 at 4:47

1 Answer 1


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 $x=3.$

  • $\begingroup$ I recall that the teacher in mit ocw calculus always said the calculus is actually the easy part. $\endgroup$
    – AlexWei
    Commented May 13, 2022 at 4:37
  • $\begingroup$ @AlexWei Not for this problem. For this problem, the easy approach was to examine the values for $x \in \{0.01,1,2,3,4\}$, and then just use intuition. This analogizes to the suggestion of DatBoi, following your posting, of manually graphing the two functions. $\endgroup$ Commented May 13, 2022 at 4:39
  • $\begingroup$ Yes. Good suggestion about mannully graphing. $\endgroup$
    – AlexWei
    Commented May 13, 2022 at 4:42
  • $\begingroup$ @AlexWei You can't really use Calculus, on something like Max$[f(x),g(x)]$, until you first nail down the interval where $f(x) < g(x)$ and the interval where $f(x) > g(x).$ However, once you do that, for this particular problem, it is game over anyway, the problem is solved. $\endgroup$ Commented May 13, 2022 at 4:42
  • $\begingroup$ The calculus part for me in this question is to notice the slope(derivative). The rest fall into the other part. $\endgroup$
    – AlexWei
    Commented May 13, 2022 at 4:44

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