Question #98b6a

We can find out whether the function is increasing or decreasing by looking at the derivative of the function. If the derivative is positive, the function is increasing, if the derivative is negative, the function is decreasing.

The way we find the intervals is by looking at the critical points (where the derivative equals zero or is undefined). The intervals between the critical points will only either be increasing or decreasing for the entire interval until the next critical point, so we can find the intervals where the function decreases or increases.

Let's start by computing the derivative:
#d/dx((e^x+2)/(e^x-1))#

We can use the quotient rule to get:
#=(d/dx(e^x+2)*(e^x-1)-(e^x+2)*d/dx(e^x-1))/(e^x-1)^2#

We can then use the chain rule to evaluate the two derivatives in the numerator:
#=(e^x*(e^x-1)-(e^x+2)*e^x)/(e^x-1)^2#

If we expand the parenthesis, we get:
#(cancel(e^x)^2-e^x-cancel(e^x)^2-2e^x)/(e^x-1)^2=(-3e^x)/(e^x-1)^2#

Now all we need to do is find the critical points of this:

First, we'll try to solve for any #x# where the function is #0#:
#(-3e^x)/(e^x-1)^2=0#

We can then multiply both sides by #(e^x-1)^2#:
#(-3e^x)/cancel((e^x-1)^2)cancel((e^x-1)^2)=0*(e^x-1)^2#

#-3e^x=0#

#e^x=0/-3#

#e^x=0#

#x=ln(0)#

Because #ln(0)# is undefined, we can conclude that the derivative is never equal to #0#.

Now we need to look at when the function is undefined. Because dividing by #0# makes the expression undefined, we want to look at when the denominator equals #0#:
#(e^x-1)^2=0#

#e^x-1=sqrt0#

#e^x=1#

#x=ln(1)=0#

So we know #x=0# is a critical point. We therefor need to evaluate whether the function is increasing or decreasing on the intervals #(-oo, 0)# and #(0, oo)#. To do this, we just pick a point in either of the intervals and evaluate the derivative to see if it's positive.

For #(-oo, 0)#, I'm going to pick #-1#:
#(-3e^-1)/(e^-1-1)^2~=-2.76#

This is negative, so we can conclude that the function is decreasing on the interval #(-oo, 0)#.

Now I'm going to pick a point in the interval #(0, oo)#. I'll use #1#:
#(-3e^1)/(e^1-1)^2~=-2.76#

This value is also negative, so we know that the function is also decreasing on the interval #(0, oo)#