Applications of Derivatives Question - None

Question

Let $f:(0,\infty) \to R$ be a differentiable function satisfying $f(x) + e^{f(x)} = \dfrac{2}{x} - \ln x - 1$. Find the number of integers in the range of $x$ satisfying the inequality $f(2x^2+1) - f(x^2+5) \ge f(1),\, x > 0$.

Accepted Answer

Step 1: Let g(x) = f(x) + e^(f(x)) = 2/x - ln(x) - 1. Differentiate both sides:

g'(x) = f'(x) + e^(f(x))·f'(x) = f'(x)[1 + e^(f(x))]

Also, g'(x) = -2/x² - 1/x < 0 for all x > 0

Step 2: Since 1 + e^(f(x)) > 0 always, and g'(x) < 0, we have f'(x) < 0. Thus f is strictly decreasing on (0,∞).

Step 3: The inequality f(2x²+1) - f(x²+5) ≥ f(1) can be rewritten as:

f(2x²+1) - f(1) ≥ f(x²+5) - f(1)

Step 4: Since f is strictly decreasing: f(a) - f(1) ≥ f(b) - f(1) is equivalent to f(a) ≥ f(b), which means a ≤ b.

Step 5: Therefore: 2x²+1 ≤ x²+5

x² ≤ 4

-2 ≤ x ≤ 2

Step 6: Since x > 0, we have 0 < x ≤ 2.

The integers in this range are: x ∈ {1, 2}

∴ Answer: 2

Question:

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