Applications of Derivatives
Increasing and Decreasing Functions
Grade 12
Question:
<p>Let <i>f</i>(<i>x</i>) = 4 tan <i>x</i> − tan² <i>x</i> + tan³ <i>x</i>, where <i>x</i> ≠ <i>n</i>π + π/2, <i>n</i> ∈ ℤ, then</p>
<p>(a) <i>f</i>(<i>x</i>) is increasing for all <i>x</i> ∈ ℝ</p>
<p>(b) <i>f</i>(<i>x</i>) is decreasing for all <i>x</i> ∈ ℝ</p>
<p>(c) <i>f</i>(<i>x</i>) is increasing in its domain</p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: To determine monotonicity of f(x), we need to find f'(x) and analyze its sign throughout the domain. A function is increasing on its domain if f'(x) > 0 for all x in the domain.
<p><strong>Step 1: Find f'(x)</strong></p><p>Given: f(x) = 4tan x − tan²x + tan³x</p><p>Let t = tan x, so f = 4t − t² + t³</p><p>f'(x) = (4 − 2tan x + 3tan²x) · sec²x</p><p>f'(x) = sec²x(4 − 2tan x + 3tan²x)</p><p><strong>Step 2: Analyze the sign of f'(x)</strong></p><p>Since sec²x = 1 + tan²x > 0 for all x in the domain, the sign of f'(x) depends on g(t) = 4 − 2t + 3t², where t = tan x.</p><p><strong>Step 3: Check if g(t) = 3t² − 2t + 4 > 0</strong></p><p>For the quadratic 3t² − 2t + 4:</p><p>Discriminant Δ = (−2)² − 4(3)(4) = 4 − 48 = −44 < 0</p><p>Since Δ < 0 and the leading coefficient 3 > 0, we have g(t) > 0 for all real t.</p><p><strong>Step 4: Conclusion</strong></p><p>Since sec²x > 0 and (4 − 2tan x + 3tan²x) > 0 for all x in the domain of f:</p><p>f'(x) > 0 for all x in the domain of f</p><p>Therefore, f(x) is strictly increasing on each interval of its domain (which consists of intervals excluding points x = nπ + π/2).</p><p>The correct interpretation is that f(x) is increasing in its domain, which accounts for the domain restrictions.</p><p><strong>∴ Answer: c</strong></p>
Correct Answer: c