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?
Best Answer
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.