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Define $f(x)=1+\dfrac{1}{x}$ on the interval $[0,+β)$. How can you find the absolute extrema of the function on this interval?
My first step was to take the first derivative, which gave me $-\dfrac{1}{x^2}$. I set it equal to $0$ and found no value of x that satisfied it.
Now all that's left is the endpoints. I know that $\lim_{x\to0^+}f(x)=β$, but does it count? Also, $\lim_{x\toβ}f(x)=1$, but again, does it count?
The function is not defined at $[0,\infty)$ because it is not defined at $0$. The fact that the limit goes to infinity as $x$ goes to $0$ means there is no maximum. As $x$ goes to positive infinity $f$ goes to $1$ but since $f$ never takes on the value of $1$, it has no minimum either.
The absolute extrema on an interval $I$, if it exists, is the number $M\in\mathbb R$ that satisfies $\forall x\in I,\,f(x)\le M$ and $\exists x_0\in I,\,f(x_0)=M$ (in other words $M=\max\{f(x)\,\mid\,x\in I\}$).
In your case $I=(0,\,+\infty)$ (the function isn't defined at $0$). We have $\forall x\in I,\,f'(x)=-\frac{1}{x^2}<0$. Thus the function is decreasing. Wouldn't this mean that the extrema is the minima of $I$? But $I$ has no minima so the function can't have an extremum.
You can also show, using $\lim\limits_{x\to 0^+}f(x)=+\infty$, that $f$ has no supremum on $I$.
A map $f: \mathbb{R} \to \mathbb{R}$ is said to have an extremum at a point $a \in \mathbb{R}$ iff there is some open ball $V^{a}$ of center $a$ such that either $f(a) \geq f(x)$ for all $x \in V^{a}$ or $f(a) \leq f(x)$ for all $x \in V^{a}$. Morevoer, it can be shown that $a$ is an extremum of $f$ only if $f'(a) = 0$.
The map $f: x \mapsto 1/x$ carries $]0, \infty[$ to $\mathbb{R}$ and is not defined at $x=0$. As you have noticed, there is no $x > 0$ such that $f'(x) = -1/x^{2} = 0$; hence $f$ has no extremum.
A problems regarding absolute extrema is going around critical points and end points (end interval of function). If you can't find the critical points then just check from both end point.
But just like john mentioned, the interval of the function doesn't allowed one to obtain a single exact point so there is no absolute maximum or minimum here and since critical points are also unavailable.
In fact, this function does not even have a single relative extrema as the graph just start at infinitesimal close to unity and go up into infinity, and remember infinity is NOT a number it is a CONCEPT.
