Applications of Derivatives
Monotonicity
Grade 12
Question:
<p>For \(f(x) = 2x - \tan^{-1} x - \ln(x + \sqrt{1+x^2})\), determine the monotonicity of \(f\).</p>
<p>(a) strictly increases \(\forall x \in \mathbb{R}\)</p>
<p>(b) strictly increases only in \((0, \infty)\)</p>
<p>(c) strictly decreases \(\forall x \in \mathbb{R}\)</p>
<p>(d) strictly decreases in \((0, \infty)\) and strictly increases in \((-\infty, 0)\)</p>
Step-by-Step Solution
Key Concept: Differentiate carefully and show that $f'(x) < 0$ for all real $x$.
<p><strong>Solution:</strong> Compute the derivative:</p><p>$$f'(x) = 2 - \frac{1}{1+x^2} - \frac{1}{\sqrt{1+x^2}}$$</p><p>Note that $\frac{d}{dx}\ln(x + \sqrt{1+x^2}) = \frac{1}{\sqrt{1+x^2}}$.</p><p>$$f'(x) = 2 - \frac{1}{1+x^2} - \frac{1}{\sqrt{1+x^2}} = \frac{2(1+x^2) - 1 - (1+x^2)}{(1+x^2)} - \frac{1}{\sqrt{1+x^2}}$$</p><p>Since $\frac{1}{1+x^2} < 1$ and $\frac{1}{\sqrt{1+x^2}} < 1$ for all $x$, we have $f'(x) < 0$ for all $x \in \mathbb{R}$. Thus $f$ strictly decreases.</p>
Correct Answer: c