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So we have the following function:

$f: (-5,-1) \cup (-1,5] \to \mathbb{R}$ given by

$$f(x) = \begin{cases} \frac{x^2+3}{x+1} & -5 < x < 3, x \neq 1 \\ 3(x-4)^4 & 3 \leq x \leq 5 \end{cases} $$

We want to find the extreme values of this function:

  • where they're located
  • whether it's a minimum or maximum
  • if it's a global or local minimum/maximum

So I always want to do these kind of problems as systematically as possible, so what I did first is find the derivatives of both the "pieces" and set them equal to zero. This gives us the following extreme values:

  • $(1,2)$ which is a minimum because $f"(1) >0$

  • $(-3,-6)$ which is a maximum because $f"(-3) < 0$

  • $(4,0)$ which is a minimum because $f"(4) > 0$

So with this we can also conclude that all of these are local and not global because the $y$ value of the minima is larger than the $y$ value of the maximum.

Now we look at the endpoints and find that $f(3) = 3$ and $f(5) = 3$. But I'm wondering, are these then both considered global maxima? And of course we also have $f(-5) \stackrel{?}{=} -7$, but $5$ is not really included in the domain so can we say that $5$ is the global minimum?

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1 Answer 1

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so you have global maximum at $f(5)$.

what you have at $f(-5)$ is not a minimum since it is not in the domain, so you call it infimum and this is the reason you have no global minimum.

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  • $\begingroup$ I made a mistake in the initial post, actually $f(3) = f(5) = 3$. So do you have two global maxima? That would seem weird to me $\endgroup$
    – USS Limitz
    Commented Nov 1, 2015 at 18:24
  • $\begingroup$ yes, two global maxima, why not, they are equal after all $\endgroup$ Commented Nov 1, 2015 at 18:35
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    $\begingroup$ Thanks, I was in doubt because I didn't know if a global maximum had to be unique $\endgroup$
    – USS Limitz
    Commented Nov 1, 2015 at 18:37

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