# Using closed interval method to find global maximum and minimum values of $f(x)= \sqrt{4-x^2}$

Use closed interval method to find global maximum and minimum values of $$f(x)= \sqrt{4-x^2}$$ in the interval $$[-2,1]$$

$$f'(x) = \frac{-x}{\sqrt{4-x^2}} = 0$$

Solving for critical points: $$x = 0 , 2, -2$$

Using closed interval method, I need to compute the critical values and intervals, for example, $$f(0), f(2), f(-2), f(1)$$ and the largest value is the maximum value in the interval and vice versa.

However, I was told that $$f(2)$$ is not required because at $$x=2$$ it is a singular point so it cant be differentiable. and from the graph, $$x=0$$ is the only stationary point, and I realised that $$x=-2$$ is also a singular point. So why do I need to compute $$x=-2$$ and not $$x=2$$ ? Is the reason because $$x=-2$$ is in the given interval? but still it is not differentiable, why must I take $$x=-2$$ into consideration when finding the function's global maximum and minimum?

• $f(2)$ need not be considered because you are looking at the interval $[-2,1]$ and $2$ is not in that interval. $f(-2)$ and $f(1)$ should be considered as this is a closed interval and you should always look at the end points. And $f(0)$ because $f'(0)=0$ Commented Oct 14, 2022 at 7:29
• @henry oh, I did not pay attention to the intervals! Commented Oct 14, 2022 at 7:34
• note: $y=f(x)$ then $y\ge 0$ and $x^2+y^2=4$ so this is the upper semi-circle of radius $2$.
– zwim
Commented Oct 14, 2022 at 9:35
• @zwim - only part of that semicircle, due to the restricted interval Commented Oct 14, 2022 at 13:29

1. we have $$f(-2)=0$$ and $$f(x) \ge 0$$ for all $$x \in [-2,1]$$, hence $$\min \{f(x): x \in [-2,1]\}= f(-2)=0.$$
2. we have $$f(0)=2$$ and $$f(x) \le 2$$ for all $$x \in [-2,1]$$, hence $$\max \{f(x): x \in [-2,1]\}= f(0)=2.$$