Define $f(x)=1+\dfrac{1}{x}$ on the interval $[0,+∞)$. 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 $\lim_{x\to0^+}f(x)=∞$, but does it count? Also, $\lim_{x\to∞}f(x)=1$, but again, does it count?
On
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 $I$.
On
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.
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.