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Discovering the max/min, if $f'(x)=0$ seems to give only the other?

E.g.

$$\sqrt{4-y^2}$$

One can reason that the maximum of this occurs when $4-y^2$ is greatest. By differentiating one finds

$$(4-y^2)'=-2y := 0$$ exactly when $y=0$. By pluggin this in one discovers that it cannot be the minimum (since there are clearly smaller values), so it must be the maximum.

However, how can I discover the minimum with the derivative?

(one can of course find it by considering that $4-y^2 \ge 0$ must hold. So the minimum of this is (in this case seen to occur when $4-y^2=0$.

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The "first derivative equal to zero" criterion only works for maxima and minima in the interior of the set on which the function is defined.

In your case the function $y \mapsto \sqrt{4-y^2}$ is defined only for $y$ values between $-2$ and $2$. The graph of the function looks a bit like a cap $\bigcap$ and the minima are achieved at the boundaries and not in the interior of the interval $[-2,2]$. This is why the derivative is not zero at the minima. Note that the derivative at the maximum is still zero because it is in the interior of $[-2,2]$.

In general, if a differentiable function is defined for all real numbers you will be able to find all local maxima and minima using the "first derivative equal to zero" criterion. If the function is only defined on a subset of the real numbers you will also have to check the boundaries of that subset because the "first derivative equal to zero" criterion only works in the interior.

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The function $4-y^2$ itself does not have minimum, when the input has very large absolute value then the output also has very large absolute value and negative (in other words, limit at infinity or negative infinity is negative infinity). But the function $\sqrt{4-y^2}$ only defined on $[-2,2]$, so you can use the closed interval method: calculate value of the function at the critical points and the boundary points to find absolute max and min. The only critical point is $y=0$ as you observed. Values of the function at -2 and 2 are 0, and value at 0 is 2. So the function has absolute max 2 and absolute min 0.

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note that $$(\sqrt{1-y^2})'=\frac{1}{2}(1-y^2)^{-1/2}(-2y)$$ and for $y=0$ we get a maximum

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