# Determine the function f=ln(x+1) - a*x if the function is increasing on the interval (-1;infinite).

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f(x) = ln (x+1) - ax

The function is increasing on the interval (-1, infinite)

Then, the first derivative 'f'(x)" is positive on the interval (-1, inf.)

Let us calculate first derivative:

f'(x) = 1/(x+1) - a

= [1-a(x+1)]/(x+1)

= (1-ax -a)/(x+1)

In order for the function to be increasing, then f'(x)>0 , then,

1-ax -a > 0

==> -ax-a > -1

Multiply by -1:

==> ax +a < 1

==> a(x+1) < 1

==> a < 1/(x+1)

If f(x) is increasing over the interval (-1,infinite), then the first derivative of the function is positive over (-1,infinite).

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

f'(x)=1/(x+1) - a

f'(x)=(1-ax-a)/(x+1)

The denominator is strictly positive over the interval (-1,infinite). For f'(x)>0, the numerator has to be positive, too.

1-ax-a>0

-ax-a>-1

We'll multiply inequality by(-1) and the inequality orientation will be change:

ax+a<1

We'll factorize:

a(x+1)<1

We'll divide by x+1:

So, a<1/(x+1), for any value of x from the interval (-1,infinite).

f(x) = ln(x+1) -a*x.

Given that ln(x+1)- ax is increasing in the interval (-1; infinity).

Since the function is increasing in (-1 , inf),

ln(x+1) -ax > 0 in (-1,inf) iff {ln(1+x) - ax }' > 0

{ln(1+x)}' - (ax)' > 0.

1/(1+x) - a > 0

1/(1+x) > a.

1 > a(x+1), Is valid as 1+x is positive for x > -1.

1 > ax +a.

1-a > ax

1/a -1 > x which implies 1/a should be ifinite.

Or a = 0.

When a = 0, ln(1+x) is an increasing function.