Question

Find the interval(s) in which the function $f(x) = \frac{x}{\log x}$, where $x \in (0, 1) \cup (1, \infty)$, is increasing.

Hint:

When working with rational functions where the denominator is squared, you can completely ignore the denominator when testing for the inequality sign, provided you ensure the points where the denominator becomes zero are omitted.

Answer 1

Concept: For a differentiable function $f(x)$ to be strictly increasing within an interval, its first derivative with respect to $x$ must be strictly positive ($f'(x) > 0$) throughout that interval.
• Quotient Rule of Differentiation: If $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}$.
• Logarithm Properties: $\log x$ refers to the natural logarithm $\ln x$ base $e$. Remember that $\log x < 0$ when $0 < x < 1$ and $\log x > 0$ when $x > 1$.

Step 1: Compute the first derivative $f'(x)$ using the quotient rule.
The given function is: \[ f(x) = \frac{x}{\log x} \] Applying the quotient rule with $u = x$ and $v = \log x$: \[ f'(x) = \frac{(\log x) \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(\log x)}{(\log x)^2} \] Since $\frac{d}{dx}(x) = 1$ and $\frac{d}{dx}(\log x) = \frac{1}{x}$, substituting these derivatives yields: \[ f'(x) = \frac{(\log x)(1) - x \cdot \left(\frac{1}{x}\right)}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2} \]

Step 2: Set up the inequality $f'(x) > 0$ and solve for $x$.
For the function to be increasing, we require: \[ f'(x) > 0 \quad \implies \quad \frac{\log x - 1}{(\log x)^2} > 0 \] The denominator $(\log x)^2$ is a squared term, meaning it is strictly positive for all $x$ in the domain $(0, 1) \cup (1, \infty)$. Therefore, the sign of the fraction depends completely on the numerator: \[ \log x - 1 > 0 \] \[ \log x > 1 \] Taking the exponential on both sides (with base $e$): \[ x > e^1 \quad \implies \quad x > e \] Thus, the function is strictly increasing in the interval $(e, \infty)$.