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Given a function, $f(x)=\sqrt{4x-x^2}$, what does it mean to find its critical points?

Assuming that the domain of the function in question is not explicitly stated, am I right to say that this function is well-defined only when $0 \leq x \leq 4$?

If that is the case, are points $x=0, x=2, x=4$ considered the critical points of $f$? Or is the critical point just at $x=2$?

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  • $\begingroup$ A "critical point" is a point in the domain of function, say $x_0$ while "critical value", is the value of the function at the critical point ie, $f(x_0)$. $\endgroup$
    – cosmo5
    Commented Oct 5, 2020 at 19:02
  • $\begingroup$ Sorry I meant to say "find its critical points". I'll edit the question accordingly $\endgroup$
    – Naja
    Commented Oct 5, 2020 at 19:09
  • $\begingroup$ Minor point: assuming that the square root of a negative number is undefined, then the function is well defined $\iff 0 \leq (4x - x^2) = x(4-x).$ Let $g(x) = [x(4-x)]$. For $x \geq 0, g(x) \geq 0 \iff 4 \geq x.$ For $x < 0$, the first factor of $g(x)$ must be negative, and the 2nd factor must be positive. Therefore, if $x < 0$, then $g(x)$ must be negative. This agrees with your assertion; I simply wasn't sure how you arrived at your assertion. $\endgroup$ Commented Oct 5, 2020 at 19:18

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From Wikipedia:

When dealing with functions of a real variable, a critical point is a point in the domain of the function where the function is either not differentiable or the derivative is equal to zero

In your case the function is not differentiable in $x=0$ and $x=4$. On the other hand:

$$f'(x)=\frac{2-x}{\sqrt{4x-x^{2}}}$$

Therefore $f'(x)=0$ only if $x=2$, that critical point would correspond to the maximum of the function in the domain $x\in [0,4]$

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  • $\begingroup$ If thats the case, then $x=0, x=4$ shouldn't be considered as critical points because critical points are located IN the interior of the domain, and should not be located at the end points correct? If so, is it also right to say that the only local extreme value is the local maximum at $x=2$? $\endgroup$
    – Naja
    Commented Oct 5, 2020 at 19:05
  • $\begingroup$ Take into account that in order to be differentiable, a function has to be continuous, therefore the function is not differentiable at the limits of its domain (the limit when $x \rightarrow 0^{-}$ or $x\rightarrow4^{+}$ does not exist), thus the boundary of the domain is formed by critical points $\endgroup$
    – Leibniz
    Commented Oct 6, 2020 at 12:30

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