Applications of Derivatives Question - None

Question

Given two functions $F(x) = \left(1+\dfrac{1}{x}\right)^x$, $G(x) = \left(1+\dfrac{1}{x}\right)^{x+1}$ defined for all $x > 0$. Which of the following is decreasing function for $\forall\, x > 0$?

Accepted Answer

Step 1: For F(x) = (1+1/x)^x, take ln: ln F(x) = x·ln(1+1/x)

Step 2: Differentiate using product rule:

d/dx[ln F(x)] = ln(1+1/x) + x·1/(1+1/x)·(-1/x²)

= ln(1+1/x) - 1/(x+1)

Step 3: Let u = 1/x, so we analyze: ln(1+u) - u/(1+u) for u > 0

Let h(u) = ln(1+u) - u/(1+u). Then h'(u) = 1/(1+u) - 1/(1+u)² = u/(1+u)² > 0

Since h(0) = 0 and h is increasing, h(u) > 0 for all u > 0. Thus F'(x)/F(x) > 0, so F(x) is increasing.

Step 4: For G(x) = (1+1/x)^(x+1), ln G(x) = (x+1)·ln(1+1/x)

d/dx[ln G(x)] = ln(1+1/x) + (x+1)·(-1/x²)/(1+1/x)

= ln(1+1/x) - (x+1)/(x(x+1)) = ln(1+1/x) - 1/x

Step 5: Let u = 1/x again: analyze ln(1+u) - u for u > 0

Let φ(u) = ln(1+u) - u. Then φ'(u) = 1/(1+u) - 1 = -u/(1+u) < 0 for u > 0

Since φ(0) = 0 and φ is decreasing, φ(u) < 0 for all u > 0. Thus G'(x)/G(x) < 0, so G(x) is decreasing.

∴ Answer: CD (G(x) is the decreasing function)

Question:

Step-by-Step Solution

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