Which are the values of the f function f(x) = m/(x + 1) - mx, to increase on the set (-1, infinite )
f(x) = m/(x+1) -mx
To find where the function is increasing, we need to calculate the first derivative, if f'(x) is positive, then the function is increasing for the interval.
Let us calculate f'(x)
f'(x) = -m/(x+1)^2 -m
= -m[1/(x+1)^2 +1]\
For the interval (-1, inf):
The value between that parenthesis [1/(x+1)^2 + 1} is always positive for all real numbers
Then in order for f'(x) to be positive, -m will have to be a positive value, Them (m) should be a negative value.
Then f(x) is increasing on the interval (-1,inf) if and only if m<0.
f(x) = m(x+1) -mx. To find the values of the function (-1 to infinity)
Let f(x) = m/(x+1) -mx .
f'(x) = -m/(x+1)^2 -m should be > 0 for x over (-1, infinity.).
-m/(x+1)^2 > m. Or
-1/(x+1)^2 > 1. Or
-1 > (x+1)^2*1. Or
-1 > (x+1)^2. Or
But (x+1)^2 is positive quantity and so can not be a < 1. So there cannot be any value of x over(-1 infinity) for which f(x) = m/(x+1) - mx is increasing.
For the function f(x) to increase between the limits (-1,infinity), the first derivative of the function, has to be positive between the same limits.
Because we're working on the set (-1,infinity), the denominator will be strictly positive.
In order to have f'(x)>0, -m>0
We'll multiply by -1 and the inequality will change: