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If I have the function $\ f(x) = x^2 + x - 2\ $ defined when $\ -5 \le x \le 10$, then we have $f'(x) = 2x + 1\ $ and $\ f''(x) = 2$.

I can easily find that there is a critical point at $x = -1/2$. It clearly is a minimum since the second derivative is positive at this $x = -1/2$. I know (since its a parabola) that this is a global minima, but how do I prove that this is global and not local?

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  • $\begingroup$ $f(x)=(x+.5)^2-2.25$ $\endgroup$ – Bill Kleinhans Sep 5 '13 at 1:57
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Note that the global minimum of a continuous function $f(x)$ in a closed interval $[a,b]$ is the minimum over the following candidates:

  • Evaluating $f$ at the endpoints: $f(a),f(b)$.
  • Evaluating $f$ at the critical points: $f(c)$, where $f'(c)=0$ or $f$ is not differentiable at $c$.

Hence, since $f(-5) = 18$ and $f(10) = 108$ are both larger than $f(-1/2) = -9/4$, it follows that the local minimum at $x=-1/2$ is also a global minimum, as desired.

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    $\begingroup$ A continuous function you mean.. $\endgroup$ – nbubis Sep 5 '13 at 0:22
  • $\begingroup$ @nbubis Yes, good catch. $\endgroup$ – Adriano Sep 5 '13 at 0:42
  • $\begingroup$ @Adriano how to recognise if it is local minima $\endgroup$ – Tesla Dec 29 '15 at 3:57
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    $\begingroup$ Maybe you also need to add differentiable in there, or at least add $f'(c) =$ undefined"? If the function has a cusp I think it could be an extreme $\endgroup$ – Ovi Oct 23 '16 at 3:47

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