<p>Let <em>f</em> and <em>g</em> be two differentiable functions on <em>R</em> such that <em>f</em>'(<em>x</em>) > 0 and <em>g</em>'(<em>x</em>) > 0 for all <em>x</em> ∈ <em>R</em>. Then for all <em>x</em>:</p><p>Which of the following is true?</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>Both (1) and (3)</p>
Step-by-Step Solution
Key Concept: When both f and g are strictly increasing (f' > 0, g' > 0), their composition and sum preserve monotonicity. Use the chain rule: if f and g are both increasing, then f(g(x)) is increasing because d/dx[f(g(x))] = f'(g(x))·g'(x) > 0 (product of positive terms).
<p><strong>Step 1:</strong> Given f'(x) > 0 and g'(x) > 0 for all x ∈ ℝ, both f and g are strictly increasing functions.</p><p><strong>Step 2:</strong> For composition f(g(x)): d/dx[f(g(x))] = f'(g(x))·g'(x). Since f'(g(x)) > 0 and g'(x) > 0, their product is positive, so f∘g is strictly increasing.</p><p><strong>Step 3:</strong> Similarly, for g(f(x)): d/dx[g(f(x))] = g'(f(x))·f'(x) > 0, so g∘f is also strictly increasing.</p><p><strong>Step 4:</strong> For f(x) + g(x): d/dx[f(x) + g(x)] = f'(x) + g'(x) > 0, so the sum is strictly increasing.</p><p><strong>Step 5:</strong> However, f(g(x)) versus g(f(x)) cannot be compared universally—their relationship depends on the specific functions. The only universally true statement is that both f(g(x)) and g(f(x)) are individually strictly increasing functions.</p><p>∴ Answer: D (The option stating that f∘g, g∘f, and f+g are all strictly increasing)</p>
Correct Answer: D