Q.Find the intervals in which the function $f$ given by $f(x) = x^3 + \frac{1}{x^3}, x \neq 0$ is

Concept: Monotonic Function Analysis — we find $f'(x)$ and check its sign.

Step 1: Differentiate

$f(x) = x^3 + x^{-3}$

$f'(x) = 3x^2 - 3x^{-4} = 3\left(x^2 - \frac{1}{x^4}\right) = \frac{3(x^6 - 1)}{x^4}$

Step 2: Sign analysis

Denominator $x^4 > 0$ for all $x \neq 0$, so sign of $f'(x)$ = sign of $x^6 - 1$.

$x^6 - 1 > 0 \implies |x| > 1$

$x^6 - 1 < 0 \implies |x| < 1$ (and $x \neq 0$)

Step 3: Intervals …

To decide where a function increases or decreases, we look at its derivative. If $f'(x) > 0$, the function is increasing; if $f'(x) < 0$, it is decreasing. The trick here is that $f$ is not defined at $x = 0$, so we must treat that point separately.

Let’s find $f'(x)$.

1. Differentiate $f(x) = x^3 + x^{-3}$. Using the power rule:

2. Factor the derivative Write everything over a common denominator:

Since $x^4 > 0$ for all $x \neq 0$, the sign of $f'(x)$ is entirely determined by the numerator $3(x^6 - 1)$. The factor $3$ is positive, so we only need to check where $x^6 - 1$ is positive or negative.

3. Solve $x^6 - 1 = 0$

   $x^6 = 1$ gives $x = \pm 1$ (real solutions). These are the critical points where $f'(x) = 0$.

4. Sign analysis of $x^6 - 1$

   For $x^6$: when $|x| < 1$, $x^6 < 1$; when $|x| > 1$, $x^6 > 1$.

   So:

   • If $x < -1$ or $x > 1$, then $x^6 > 1 \Rightarrow x^6 - 1 > 0 \Rightarrow f'(x) > 0$ → increasing.
   • If $-1 < x < 1$ (and $x \neq 0$), then $x^6 < 1 \Rightarrow x^6 - 1 < 0 \Rightarrow f'(x) < 0$ → decreasing.
   Watch out

   Do not forget that $x = 0$ is excluded from the domain. The function is not defined there, so we split the interval $(-1, 1)$ into $(-1, 0)$ and $(0, 1)$. Both are decreasing.

5. Check the critical points $x = \pm 1$ …

Method: Monotonicity of a Function with a Domain Exclusion

This method applies to increasing/decreasing questions where the function is undefined at one or more points (like $x=0$ here), which forces the intervals to be split at those excluded points, not just at the zeros of the derivative.

Steps

Step 1: Differentiate using the power rule

Rewrite negative-power terms as $x^{-n}$ before differentiating, e.g. $x^3+x^{-3}$, so the power rule applies directly to each term.

Step 2: Combine into a single fraction and factor

Put $f'(x)$ over a common denominator (usually the lowest power of $x$ appearing) so the sign of $f'(x)$ reduces to the sign of a single polynomial numerator divided by a denominator whose sign is easy to determine (an even power of $x$ is always non-negative).

Step 3: Solve for where the numerator is zero

These are the critical points. Solve $x^k = c$-type equations carefully, keeping only the real roots. …

Common Mistakes

Mistake 1: Treating the domain-exclusion point as just another zero of $f'$

Why it's wrong: $x=0$ is not a critical point here (the function isn't even defined there) — it is a genuine hole in the domain, and the interval $(-1,1)$ must be reported as the split pieces $(-1,0)$ and $(0,1)$ rather than left as one interval, even though $f'$ has the same sign on both pieces. Correct approach: always list domain-exclusion points separately from zeros of $f'$ when building the interval breakdown, and report them as split intervals.

Mistake 2: Forgetting that an even power never changes sign …