Q.Let I be any interval disjoint from $[-1, 1]$. Prove that the function $f$ given by $f(x) = x + \frac{1}{x}$ is increasing on I.

Concept: Monotonic Function Analysis — we check the sign of the derivative on the given interval.

Step 1: Compute the derivative:

Step 2: The denominator $x^2 > 0$ for all $x \neq 0$, so the sign of $f'(x)$ is the sign of $x^2 - 1$.

Step 3: Since $I$ is disjoint from $[-1, 1]$, every $x \in I$ satisfies $|x| > 1$. Hence $x^2 > 1$, so $x^2 - 1 > 0$. …

We need to show that $f(x) = x + \frac{1}{x}$ is increasing on any interval $I$ that has no overlap with $[-1, 1]$. That means every point in $I$ satisfies either $x < -1$ or $x > 1$.

The standard tool for proving a function is increasing on an interval is the derivative test: if $f'(x) > 0$ for all $x$ in the interval (and $f$ is continuous), then $f$ is strictly increasing there. So let's compute the derivative.

This is defined for all $x \neq 0$, which is fine since our interval $I$ avoids $[-1, 1]$ and therefore avoids $0$ as well.

Now, when is $f'(x) > 0$? Solve:

This means $|x| > 1$, i.e., $x < -1$ or $x > 1$. Exactly the condition that defines $I$.

So on any interval $I$ disjoint from $[-1, 1]$, every point satisfies $|x| > 1$, hence $f'(x) > 0$ everywhere on $I$. Since $f$ is differentiable (and therefore continuous) on $I$, the derivative being positive throughout implies $f$ is strictly increasing on $I$. …

Method: Proving Monotonicity on an Interval Described as "Disjoint From" a Given Set

Use this method when the interval isn't given explicitly as numbers, but is instead described in words, such as "any interval disjoint from $[a,b]$" — you must translate that description into an algebraic condition before you can use it.

Steps

Step 1: Differentiate the function and simplify to a single fraction if possible.

Combining terms over a common denominator (e.g. $f'(x)=1-\dfrac{1}{x^2}=\dfrac{x^2-1}{x^2}$) makes the sign much easier to read off, since a squared denominator is always positive wherever defined.

Step 2: Translate the verbal description of the interval into an algebraic inequality on $x$.

"Disjoint from $[-1,1]$" means every point of the interval satisfies $x<-1$ or $x>1$, i.e. $|x|>1$, i.e. $x^2>1$.

Step 3: Match this algebraic condition directly against the sign condition needed for the derivative. …

Common Mistakes

Mistake 1: Picking one specific numeric interval and proving the result only for that example.

Why it's wrong: the question asks to prove the claim for ANY interval disjoint from $[-1,1]$, so a proof tied to one numeric example doesn't establish the general statement. Correct approach: keep the argument in terms of the algebraic condition $x^2>1$, which covers every such interval at once.

Mistake 2: Not noticing that the disjointness condition already excludes $x=0$, the one point where the derivative formula is undefined. …