calculusoptimization
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?
We easily get $f(x,y)\ge 0$ and $$f(x,y)=0\iff x^2=4\;\wedge\;y=0\iff (x,y)=(-2,0)\;\vee\;(x,y)=(2,0).$$ Obviously $f(\pm 2,0)=0$, so local minima are also global.
To answer this question one has to review the definitions. Specialized to this case these definitions read: A function $f: \mathbb R \to \mathbb R$ has a global maximum point at $x^∗ \in \mathbb R$ if $f(x^∗) \geq f(x)$ for all $x$ in $\mathbb R$. Similarly, $f$ has a global minimum point at $x^∗ \in \mathbb R$ if $f(x^∗) \leq f(x)$ for all $x$ in $\mathbb R$.
Clearly, for the given function $f$ the following holds: for every $x^∗ \in \mathbb R$ there exists $x \in \mathbb R$ such that $f(x^*) \lt f(x)$. This is the negation of the definition of the global maximum. So, $f$ does not have a global maximum. Similarly, $f$ does not have a global minimum.
A function $f : \mathbb R \to \mathbb R$ is said to have a local maximum point at the point $x^∗ \in \mathbb R$ if there exists some $\epsilon \gt 0$ such that $f(x^∗) \geq f(x)$ for all $x \in (x^*-\epsilon, x^*+\epsilon)$. Similarly, $f : \mathbb R \to \mathbb R$ is said to have a local minimum point at the point $x^∗ \in \mathbb R$ if there exists some $\epsilon \gt 0$ such that $f(x^∗) \leq f(x)$ for all $x \in (x^*-\epsilon, x^*+\epsilon)$.
It is clear that for $\epsilon = 1/2$ we have $f(0) \leq f(x)$ for all $x \in (-1/2,1/2)$. Therefore, $f$ has a local minimum at $x^* = 0$.
It is a good exercise to state the negation of the statement: "$f$ has a local maximum" and prove it for the given $f$.
Finally, the correct answer is (a).
A graph is always helpful:
Best Answer
Note that the global minimum of a continuous function $f(x)$ in a closed interval $[a,b]$ is the minimum over the following candidates:
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.