giorgiana1976 | Student

We'll note the given expression:

f(x) = lnx-2(x-1)/(x+1)

We'll have to demonstrate that f(x)>0.

To prove that a function is increasing, we'll have to prove that the first derivative of the function is positive.

We'll calculate the first derivative:

f'(x) = [lnx-2(x-1)/(x+1)]'

The ratio from the expression of the function will be differentiated using the quotient rule.

f'(x) = (1/x)-{[2(x-1)'*(x+1)-2(x-1)*(x+1)']/(x+1)^2}

f'(x) = (1/x)-(2x+2-2x+2)/(x+1)^2

f'(x) = (1/x)-(4)/(x+1)^2

f'(x) = [(x+1)^2-4x]/x*(x+1)^2

f'(x) = (x^2+2x+1-4x)/x*(x+1)^2

f'(x) = (x^2-2x+1)/x*(x+1)^2

f'(x) = (x-1)^2/x*(x+1)^2

We notice that for any x>1, (x-1)^2>0 and x*(x+1)^2>0, so f(x)>0.

f(x) = lnx-2(x-1)/(x+1)

lnx-2(x-1)/(x+1) > 0 q.e.d.

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