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Question
Show that y = `log(1+x) - (2x)/(2+x), x> - 1`, is an increasing function of x throughout its domain.
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Solution
We have,
y = log `(1 + x) - (2x) / (2 + x), x > -1`
Here, `dy/dx = 1/ (1 + x) - 2 d/dx (x/ (2 +x))`
`1/ (1 + x) - 2 {(2 + x) * 1- x (0 +1)}/(2 + x)^2`
`1/ (1 +x) - 4/ (2 + x)^2 = ((2 + x)^2 - 4 (1 + x))/((1 + x) (2 + x)^2)`
`= x^2/((1 + x) (2 + x)^2) AA x > - 1`
x2 > 0, (2 + x)2 >0 (being perfect square) and (1 + x) > 0 ∀ x> -1
`dy/dx>= 0` for all x > -1
Hence, y is an increasing function of x throughout its domain.
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