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Question
Show that y = `log (1 + x) – (2x)/(2 + x), x > - 1` is an increasing function on its domain.
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Solution
y = `log (1 + x) – (2x)/(2 + x), x > - 1`
∴ `dy/dx =d/dx[log (1 + x) - (2x)/(2 + x)]`
= `(1)/(1 + x).d/dx(1 + x) - ((2 + x).d/dx(2x) - 2x.d/dx(2 + x))/(2 + x)^2`
= `(1)/(1 + x) xx (0 + 1) ((2 + x) xx 2 - 2x(0 + 1))/(2 + x)^2`
= `(1)/(1 + x) - (4 + 2x - 2x)/(2 + x)^2`
= `(1)/(1 + x) - (4)/(2 + x)^2`
= `((2 + x)^2 - 4(1 + x))/((1 + x)(2 + x)^2)`
= `(4 + 4x + x^2 - 4 - 4x)/((1 + x)(2 + x)^2`
= `(x^2)/((1 + x)(2 + x)^2) >` 0 for all x > – 1
Hence, the given function is increasing function on its domain.
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