# Alternate approach for finding Minimum value of the below function

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 ?

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

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 $$x=3.$$

• I recall that the teacher in mit ocw calculus always said the calculus is actually the easy part. Commented May 13, 2022 at 4:37
• @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. Commented May 13, 2022 at 4:39
• Yes. Good suggestion about mannully graphing. Commented May 13, 2022 at 4:42
• @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. Commented May 13, 2022 at 4:42
• The calculus part for me in this question is to notice the slope(derivative). The rest fall into the other part. Commented May 13, 2022 at 4:44