Is (1,1) a local maximum of $f$?

Question

For a function

$$ f(x)=x-6\sqrt{x-1}, $$

can $x=1$ be considered a critical number, and is $f(1)$ a local maximum?

The definition my textbook gives me of a critical number is a number that is an interior point of the domain of $f$ where $f'$ is either zero or undefined. If $f$ is continuous in $[1,\infty)$, can $x=1$ be considered as an interior point?

Answer 1

First, a definition:

Definition: A point $x$ is an interior point of a set $E \subseteq \mathbb{R}$ if there exists some $r > 0$ such that $$ B(x,r) = \{ y : |x-y| < r \} \subseteq E. $$

That is, a point is in the interior of $E$ if that point is contained in an open ball which is entirely contained within $E$.

The natural domain of $f$, defined by the formula $$ f(x) = x - 6\sqrt{x-1},$$ is the interval $[1,\infty)$ (this is the largest set of real numbers such that the formula defining $f$ gives a real number). Note that any ball in $\mathbb{R}$ which contains $x = 1$ must also contain some point $y=1-\varepsilon$, where $\varepsilon > 0$. Hence there is no $r > 0$ such that $$ B(1,r) \subseteq [1,\infty). $$ As such, $x=1$ is not an interior point of the domain of $f$. Therefore, by the definition provided, $x=1$ is not a critical point of $f$.

However, that does not mean that $f(1)$ is not a (local) maximum of $f$.

Definition: Let $f : D \to \mathbb{R}$ be a function (that is, $f$ is a real-valued function with domain $D$). A real number $M$ is a local maximum of $f$ if

1. there is some $c \in D$ such that $f(c) = M$, and
2. there is some $r > 0$ such that $f(c) \ge f(x)$ for all $x \in D$ with $|x-c| < r$.

It can be shown directly that there is an interval $[1, b)$ on which $f(x) \le 1$, which is sufficient to show that $f$ has a local maximum at $x=1$. It is instructive to work this this out from the definition, rather than appealing to theorems.

Again, the goal is to show that $f(x) \le 1$ on some (open) interval containing $1$. Doing some basic algebra, \begin{align} f(x) \le 1 &\iff x - 6\sqrt{x-1} \le 1 \\ &\iff (x-1) - 6\sqrt{x-1} \le 0 \\ &\iff \sqrt{x-1} \left( \sqrt{x-1} - 6 \right). \end{align} Observe that $\sqrt{x-1} \ge 0$ for all $x \ge 1$, and that $$ \sqrt{x-1} - 6 < 0 \iff \sqrt{x-1} < 6 \iff x-1 < 36 \iff x < 37. $$ From this, it follows that $f(x) \le 1$ on the interval $[1,37)$. Therefore $f(1) = 1$ is a local maximum of $f$.

Note that this compares favorably to a GeoGebra sketch of the problem:

Note, also, that one can appeal to theorems. Specifically, it can be shown that if $f$ is differentiable on an interval $(a,b)$ and $f'(x) < 0$ on that interval, then $f$ is decreasing. A quick computation shows that $f'(x) < 0$ on the interval $(1,10)$, hence $f$ is decreasing on that interval. It can then be inferred that $f(1)$ is a local maximum of $f$. However, there are several "gaps" in this argument, which do need to be filled in.