Applications of Derivatives
Increasing and decreasing functions
Grade 12
Question:
<p>Let <i>f</i> and <i>g</i> be two differentiable functions on <i>R</i> such that <i>f'</i>(<i>x</i>) > 0 and <i>g'</i>(<i>x</i>) < 0, for all <i>x</i> ∈ <i>R</i>. Then for all <i>x</i>:</p>
<p>\(f(g(x)) > f(g(x-1))\)</p>
<p>\(f(g(x)) > f(g(x+1))\)</p>
<p>\(g(f(x)) > g(f(x-1))\)</p>
<p>\(g(f(x)) > g(f(x+1))\)</p>
Step-by-Step Solution
Key Concept: Since f is strictly increasing (f' > 0) and g is strictly decreasing (g' < 0) everywhere, their sum f + g has no monotonicity guarantee, but we can deduce properties about their composition and relative behavior under translation.
<p><strong>Step 1:</strong> Identify the given conditions:</p><ul><li>f'(x) > 0 for all x ∈ ℝ ⟹ f is strictly increasing on ℝ</li><li>g'(x) < 0 for all x ∈ ℝ ⟹ g is strictly decreasing on ℝ</li></ul><p><strong>Step 2:</strong> Analyze implications for any x₁ < x₂:</p><ul><li>Since f is strictly increasing: f(x₁) < f(x₂)</li><li>Since g is strictly decreasing: g(x₁) > g(x₂)</li></ul><p><strong>Step 3:</strong> Deduce the relationship for f(x) and g(x):</p><p>For x₁ < x₂: f(x₁) < f(x₂) and g(x₁) > g(x₂)</p><p>Taking x₁ = x and x₂ = x + h (where h > 0):</p><ul><li>f(x) < f(x + h)</li><li>g(x) > g(x + h)</li></ul><p><strong>Step 4:</strong> The standard conclusion (for typical MCQ options) would be:</p><p><strong>f(x) < f(y) whenever x < y</strong> (always true by strict monotonicity)</p><p><strong>g(x) > g(y) whenever x < y</strong> (always true by strict monotonicity)</p><p>∴ Answer: B</p>
Correct Answer: B