Find the intervals on which a function is increasing/decreasing?

[utility]

1. The problem statement, all variables and given/known data
F(x)=6/x-(1/1-x)

Find the intervals on which the function is increasing/decreasing?


2. Relevant equations
F(x)=6/x-(1/1-x)

F'(x)= -6/x^2 -1/(1-x)^2




3. The attempt at a solution

Critical points are x=0 and x=1

Function has a discontinuitiy at 0

Checking to the right of the critical point:

f'(2) <0

f'(-1)<0

Seems like for all values I put into the derivative I get a negative number in return. And I know from graphing the function it is increasing approxmiately on (2,infinity) and decreasing (-2,-infinity)

[Stephen Tashi]

F(x)=6/x-(1/1-x)


I know from graphing the function it is increasing approxmiately on (2,infinity)

To me, it looks like the function is approaching 0 as x approaches infinity. How did you graph it?

[utility]

To me, it looks like the function is approaching 0 as x approaches infinity. How did you graph it?

Using graph calc, I entered i as (6/x)/(1/1-x). Is the other work correct?

[Stephen Tashi]

Using graph calc, I entered i as (6/x)/(1/1-x). Is the other work correct?

Shouldn't you do addition of 6/x and 1/(1-x) instead of division?

[uart]

Why don't you solve the inequality F'(x) > 0

To solve -6/x^2 -1/(1-x)^2 > 0 you can multiply both sides by x^2 (1-x)^2, and since this quantity is always positive it wont break the inequality. (Though obvious you still need to be careful of the points x=0 and x=1).

BTW. If you do the above you will find that the LHS of the inequality is negative definite and hence there is nowhere in the domain of that function where the derivative is positive.

[Ray Vickson]

1. The problem statement, all variables and given/known data
F(x)=6/x-(1/1-x)

Find the intervals on which the function is increasing/decreasing?


2. Relevant equations
F(x)=6/x-(1/1-x)

F'(x)= -6/x^2 -1/(1-x)^2




3. The attempt at a solution

Critical points are x=0 and x=1

Function has a discontinuitiy at 0

Checking to the right of the critical point:

f'(2) <0

f'(-1)<0

Seems like for all values I put into the derivative I get a negative number in return. And I know from graphing the function it is increasing approxmiately on (2,infinity) and decreasing (-2,-infinity)

For ANY x (except 0 and 1) both terms -6/x^2 and -1/(1-x)^2 are < 0, so F'(x) < 0 everywhere where it is defined (that is, everywhere except x=0 and x=1). When I graph it I do not get what you said. Are you sure you wrote the correct function here? (The function F does increase from negative values for x slightly below x = 1 to positive values just after x = 1, but it does so by "jumping", not by increasing in a smooth way.)

RGV

[utility]

For ANY x (except 0 and 1) both terms -6/x^2 and -1/(1-x)^2 are < 0, so F'(x) < 0 everywhere where it is defined (that is, everywhere except x=0 and x=1). When I graph it I do not get what you said. Are you sure you wrote the correct function here? (The function F does increase from negative values for x slightly below x = 1 to positive values just after x = 1, but it does so by "jumping", not by increasing in a smooth way.)

RGV

I graphed F(x) and it looked to be increasing on a certain interval, however when graphing F'(x) it is certainly negative everywhere it is defined.

So it ok to say the function is decreasing everywhere it is defined?