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Theorem: $$\text{Let } U \subseteq \mathbb R. f:U \rightarrow \mathbb R \text{ has a local minimum at } x_0 \text{ if there exists } \epsilon > 0 \text{ such that } f(x_0) \leq f(x) \text{ for all } x \in B_{\epsilon}(x_0) \cap U$$

How would one use this theorem to prove that the function,

$$ h(x) = \begin{cases} 5-(x-1)^2 &\text{ if } x \in (0,3),\\ |{x-4}| & \text{ if } x \not\in (0,3). \end{cases} $$

achieves its local minimum at $x=0$?

We know that if a function is a local minimum, that the function on either side of the point will be greater than the minimum point, so I was thinking that one could prove that if you took $f(x_0 + \delta) > f(x_0)$. I don't get how one would prove that $5 - (x-1)^2 +\delta$ > $5-(x-1)^2$ is true for $\delta >0$, it just seems too trivial, and hasn't proven that the point is a minimum. (Similarly for $|{x-4}|$).

Thank you.

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    $\begingroup$ you are taking $f(x_0)+\delta$ instead of $f(x_0+\delta)$ $\endgroup$
    – Maryam
    Apr 26, 2020 at 16:33
  • $\begingroup$ Your function is not defined at $0$ nor $3$? $\endgroup$
    – Bernard
    Apr 26, 2020 at 16:33
  • $\begingroup$ @Bernard I think the notation is the way it is to make the question I was given more confusing. $\endgroup$
    – user755706
    Apr 26, 2020 at 19:22

1 Answer 1

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I assume you mean that $h:\mathbb{R}\to\mathbb{R}$ is defined as $h(x)=5-(x-1)^2$ if $x\in(0,3)$ and $h(x)=\lvert x-4\rvert$ if $x\in\mathbb{R}\setminus(0,3)$.

Pick for example $\delta=1$. We want to show that for all $x\in B_\delta (0)$ we have $h(x)\geq h(0)=4$. If $0<x< 1$, then $$h(x)=5-(x-1)^2\geq 5-(0-1)^2 = 4.$$ If on the other hand $-1< x < 0$, then $$h(x)=\lvert x-4\rvert\geq \lvert 0-4\rvert = 4.$$ Hence, $h$ obtains a local minimum at 0 with value 4.

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  • $\begingroup$ Thank you, and the first part was just a general theorem to be applied to the question indirectly. $\endgroup$
    – user755706
    Apr 26, 2020 at 19:04

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