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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?

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    $\begingroup$ $f(2)$ 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 $f'(0)=0$ $\endgroup$
    – Henry
    Commented Oct 14, 2022 at 7:29
  • $\begingroup$ @henry oh, I did not pay attention to the intervals! $\endgroup$
    – user307640
    Commented Oct 14, 2022 at 7:34
  • $\begingroup$ note: $y=f(x)$ then $y\ge 0$ and $x^2+y^2=4$ so this is the upper semi-circle of radius $2$. $\endgroup$
    – zwim
    Commented Oct 14, 2022 at 9:35
  • $\begingroup$ @zwim - only part of that semicircle, due to the restricted interval $\endgroup$
    – Henry
    Commented Oct 14, 2022 at 13:29

1 Answer 1

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Without derivatives:

  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 $\max \{f(x): x \in [-2,1]\}= f(0)=2.$

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  • $\begingroup$ Many thanks for your comment. $\endgroup$
    – Fred
    Commented Oct 14, 2022 at 9:44

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