Which are the values of the f function f(x) = m/(x + 1) - mx, to increase on the set (-1, infinite )

Expert Answers
hala718 eNotes educator| Certified Educator

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.

 

  

neela | Student

f(x) = m(x+1) -mx. To find the values of the function (-1 to infinity)

Solution:

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.

 

 

 

 

 

 

 

giorgiana1976 | Student

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.

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

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

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:

m<0