Q.The function $f(x) = \dfrac{2x^2 - 1}{x^4}$, $x > 0$, decreases in the interval ______.

Concept: Derivative Sign Analysis — to find where $f$ decreases, we check where $f'(x) < 0$.

Step 1: Simplify and differentiate.

Rewrite $f(x) = 2x^{-2} - x^{-4}$. Then

Step 2: Combine into a single fraction.

Step 3: Determine sign for $x > 0$. …

To decide where a function increases or decreases, we look at the sign of its first derivative. If $f'(x) > 0$, the function is rising; if $f'(x) < 0$, it is falling. The question asks for the interval where $f$ decreases, so we need $f'(x) < 0$.

Let’s work through it carefully.

1. Rewrite the function for easier differentiation.

   $f(x) = \dfrac{2x^2 - 1}{x^4} = \dfrac{2x^2}{x^4} - \dfrac{1}{x^4} = 2x^{-2} - x^{-4}$.

   This avoids the quotient rule and makes differentiation straightforward.

2. Differentiate term by term.

   $f'(x) = 2(-2)x^{-3} - (-4)x^{-5} = -4x^{-3} + 4x^{-5}$.

   Factor out the common factor $4x^{-5}$:

   $f'(x) = 4x^{-5}(-x^{2} + 1) = \dfrac{4(1 - x^2)}{x^5}$.

   $f'(x) = \frac{4(1 - x^2)}{x^5}$

3. Analyze the sign of $f'(x)$ for $x > 0$.

   Since $x > 0$, the denominator $x^5$ is always positive. The factor $4$ is also positive. So the sign of $f'(x)$ depends entirely on the numerator $(1 - x^2)$.

   • $1 - x^2 > 0$ when $x^2 < 1$, i.e., $0 < x < 1$. Then $f'(x) > 0$, so $f$ increases.
   • $1 - x^2 < 0$ when $x^2 > 1$, i.e., $x > 1$. Then $f'(x) < 0$, so $f$ decreases. …

Method: Sign-Analysis of the Derivative for Power/Rational Functions

When a function is built from $x$ raised to negative or fractional powers (or is a rational expression), rewriting it with exponents before differentiating avoids a messy quotient-rule computation and makes the sign analysis far cleaner.

Steps

Step 1: Rewrite the function using negative-exponent form.

Convert every $\dfrac{1}{x^n}$ term into $x^{-n}$ so ordinary power-rule differentiation applies term by term, instead of the quotient rule.

Step 2: Differentiate term by term using the power rule.

Step 3: Combine the result into a single fraction and factor the numerator.

Find a common denominator (usually the highest power of $x$ that appeared), then factor out anything common in the numerator — this exposes the derivative as (constant)$\times$(sign-determining factor)/(power of $x$).

Step 4: Note the domain restriction and the sign of the denominator. …

Common Mistakes

Mistake 1: Ignoring the given restriction $x>0$

Why it's wrong: Solving $f'(x)<0$ over all reals gives $x<-1$ or $x>1$, but the problem explicitly restricts the domain to $x>0$ — including the negative branch $x<-1$ in the final answer is a domain error, not a calculus error. Correct approach: always re-apply any stated domain restriction to the sign-analysis result before writing the final interval.

Mistake 2: Sign slip converting to negative exponents …