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3.4 · Q3

Q.Show that the function f(x)=log⁡(1+x)+11+xf(x) = \log(1 + x) + \dfrac{1}{1 + x} increases on (0,∞)(0, \infty).

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✓ Free question

f′(x)f'(x) simplifies to x(1+x)2\dfrac{x}{(1+x)^2}, which is positive for every x>0x>0; hence ff increases on (0,∞)(0,\infty).

ff is increasing on an interval if f′(x)>0f'(x)>0 throughout it.

  1. Given f(x)=log⁡(1+x)+11+xf(x)=\log(1+x)+\dfrac{1}{1+x}.
  2. Differentiate: ddxlog⁡(1+x)=11+x\dfrac{d}{dx}\log(1+x)=\dfrac{1}{1+x} and ddx(1+x)−1=−1(1+x)2\dfrac{d}{dx}(1+x)^{-1}=-\dfrac{1}{(1+x)^2}.
  3. So f′(x)=11+x−1(1+x)2f'(x)=\dfrac{1}{1+x}-\dfrac{1}{(1+x)^2}.
  4. Take LCD (1+x)2(1+x)^2: f′(x)=(1+x)−1(1+x)2=x(1+x)2f'(x)=\dfrac{(1+x)-1}{(1+x)^2}=\dfrac{x}{(1+x)^2}.
  5. For x>0x>0: numerator x>0x>0 and denominator (1+x)2>0(1+x)^2>0, so f′(x)>0f'(x)>0.
  6. Since f′(x)>0f'(x)>0 for all x∈(0,∞)x\in(0,\infty), ff is (strictly) increasing on (0,∞)(0,\infty).
✓Final answer

f′(x)=x(1+x)2>0f'(x)=\dfrac{x}{(1+x)^2}>0 for all x>0x>0; therefore ff increases on (0,∞)(0,\infty). Proved.

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