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.