Finding a min on $\sqrt{f(x)}$ is equal to min on $f(x)$ During a discussion about RMS, one said that finding the min of a function or of it square root is the same because square root is monotonic increasing.
Is this make any sense?

Finding a minimum of a function is equivalent to finding the point at which the derivative of the function is equal to zero on a given interval. If we can show that the sign of the derivative of any $f(x)$ is equal to the sign of the derivative of any $\sqrt{f(x)}$ then we can show that the min or minimums for $\sqrt{f(x)}$ exist for exactly each minimum of $f(x)$.
Case 1:
Assume $f'(x) > 0$.
$\frac{d}{dx}\sqrt{f(x)}$ by the chain rule is equal to $\frac{1}{2\sqrt{f(x)}} * f'(x)$.We know the first term must be greater than zero since the square root cannot be negative. Thus $\frac{d}{dx}\sqrt{f(x)} > 0$ 
Case 2:
Assume $f'(x) < 0$
$\frac{d}{dx}\sqrt{f(x)} = \frac{1}{2\sqrt{f(x)}} * f'(x)$ Again, the first term is positive, and the second term is assumed to be negative, yielding a negative derivative.
$\frac{d}{dx}\sqrt{f(x)} < 0$ 
Case 3:
Assume $f'(x) = 0$
$\frac{d}{dx}\sqrt{f(x)} = \frac{1}{2\sqrt{f(x)}} * f'(x)$
Thus  $\frac{d}{dx}\sqrt{f(x)} = 0$
For the last case, and the case we are after, if f'(x) = 0, the location of all possible minimums, $\frac{d}{dx}\sqrt{f(x)}$ also equals zero. Technically, we have shown that the critical points must exist at the same location,  but since we have shown also that they increase and decrease over the exact same intervals, the nature of the critical points must also match. Thus we have shown that if a function is continuous, differentiable, and strictly positive, the minimums of its square root match the minimums of the function. 
