Limits, Continuity & Differentiability
Differentiation
Grade 12
Question:
<p>If \(f(0) = 1\), \(f'(0) = -1\), \(f(x) > 0\) for all \(x\), then there exists a function \(f(x)\) such that</p>
<p>(a) \(f'(x) < 0\) for all \(x\)</p>
<p>(b) \(-1 < f'(x) < 0\) for all \(x\)</p>
<p>(c) \(-2 \leq f''(x) \leq -1\) for all \(x\)</p>
<p>(d) \(f''(x) = -1\) for all \(x\)</p>
Step-by-Step Solution
Key Concept: Use the definition of derivative at x=0 and the positivity constraint f(x)>0 to establish inequalities on the behavior of f near 0. The derivative f'(0)=-1 means f decreases sharply from f(0)=1, but must stay positive due to the given constraint.
<p><strong>Step 1:</strong> Analyze the given conditions: f(0)=1, f'(0)=-1, and f(x)>0 for all x∈ℝ.</p><p><strong>Step 2:</strong> From the definition of derivative: f'(0) = lim(h→0) [f(h)-f(0)]/h = lim(h→0) [f(h)-1]/h = -1</p><p><strong>Step 3:</strong> This means for small h>0: f(h)≈1-h (from Taylor expansion). For small positive h, f(h)≈1-h which is positive when h is small enough.</p><p><strong>Step 4:</strong> The constraint f(x)>0 for all x means f cannot cross the x-axis. A function like f(x)=e^(-x) satisfies: f(0)=1 ✓, f'(0)=-1 ✓, and e^(-x)>0 for all x ✓</p><p><strong>Step 5:</strong> We can also verify f(x)=1/(1+x) works for appropriate domain: f(0)=1 ✓, f'(x)=-1/(1+x)² so f'(0)=-1 ✓, and f(x)>0 when x>-1 ✓</p><p>∴ Answer: A</p>
Correct Answer: A