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?

  • $\begingroup$ $f(x)=(x+.5)^2-2.25$ $\endgroup$ Sep 5, 2013 at 1:57

1 Answer 1


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.

  • $\begingroup$ @Adriano how to recognise if it is local minima $\endgroup$
    – Tesla
    Dec 29, 2015 at 3:57

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