# maxima and minima piecewise function

I'm exercising on maxima and minima, I think I got the point of global and local extremes but then I find this piecewise function where my teacher says that the right answer is "c". I thought the right answer was "b". Could you help me understand why,please.

Let $f(x)= \begin{cases} 1+x, &\text{if }x\lt 0\\ 0,& \text{if }x=0\\ 1-x,&\text{if }x\gt0 \end{cases}$

Then:

a) f has a local, but not global, minimum point

b) f has a global minimum point

c) f has a local, but not global, maximum point

d) f has a global maximum point

Thank you

• The special value at x=0 makes the function discontinuous. There is clearly no global minimum because the function is not bounded from below. There is an upper bound f(x)<1, but the value 1 is not reached, as f(0)=0 instead of 1 (which would be the left/right limit). Extrema are tricky for discontinuous functions, but strictly speaking, in the neighbourhood of x=0, all the values of the function are bigger than f(0), so in some weak sense, x=0 is a local minimum. That would make (a) the correct answer. – orion Jun 7 '15 at 15:20
• @iadvd You can use \begin{cases}...\end{cases} for piecewise defined functions. (See the post.) – Martin Sleziak Jun 8 '15 at 5:57
• @MartinSleziak understood, thank you for making that change. – iadvd Jun 8 '15 at 6:03
• It might be useful to draw a picture: WolframAlpha. – Martin Sleziak Jun 8 '15 at 7:35

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).