001 - ISI Questions & Solutions
Step 1: Define a helper function Define g ( x ) = f ( x ) − x . g(x)=f(x)-x. g ( x ) = f ( x ) − x . Then g g g is differentiable on R \mathbb{R} R , and g ′ ( x ) = f ′ ( x ) − 1. g'(x)=f'(x)-1. g ′ ( x ) = f ′ ( x ) − 1. Step 2: Analyze the derivative of g g g Since ∣ f ′ ( x ) ∣ < 1 2 |f'(x)|<\tfrac12 ∣ f ′ ( x ) ∣ < 2 1 , we have f ′ ( x ) − 1 ≤ 1 2 − 1 = − 1 2 , f'(x)-1 \le \tfrac12 -1 = -\tfrac12, f ′ ( x ) − 1 ≤ 2 1 − 1 = − 2 1 , and also f ′ ( x ) − 1 ≥ − 1 2 − 1 = − 3 2 . f'(x)-1 \ge -\tfrac12 -1 = -\tfrac32. f ′ ( x ) − 1 ≥ − 2 1 − 1 = − 2 3 . Hence, g ′ ( x ) < 0 for all x ∈ R . g'(x)<0 \quad \text{for all } x\in\mathbb{R}. g ′ ( x ) < 0 for all x ∈ R . So g g g is strictly decreasing on R \mathbb{R} R . Step 3: Look at the behavior at infinity Fix any x ∈ R x\in\mathbb{R} x ∈ R . By the Mean Value Theorem applied between 0 0 0 and x x x , there exists ξ \xi ξ between 0 0 0 and x x x such that f ( ...