1st & 2nd Derivative Graphing Strategy. Needing help.

[irishpump]

I just got an exam back where I totally missed this question, but I'm redoing it in preparation for upcoming finals.

Here's the question:

For the function g(x) = x/(x+1)

When is g increasing?
When is g decreasing?
When is g concave up?
When is g concave down?
When is g' increasing?

So here's what I've done:

All real numbers except x=-1, Horizontal Asymptote=1, Vertical Asymptote=-1

g'(x)= 1/(x+1)^2 -->x+1=0 --> x=-1, Partition number= -1

Using a sign chart: g(x) 0 = g'(x) 1, g(x) -2=g'(x) 1 (-infinity, -1)=increasing and (-1, infinity)=decreasing. So g(x) is increasing from (-infinity, -1) and decreasing from (-1, infinity)

g''(x)= -2/(x+1)^3 --> Partition number: x=-1

Sign chart: g(x) 0= g''(x) -2, g(x) -2=g'' 2 So increasing from (-infinity, -1), so (-infinity, -1) is concave upward, and decreasing from (-1, infinity), so concave downward

For the final question, g' is increasing from (-infinity, -1)

Did I do this correct? I've had a harder time with this than a lot of other things that we've learned that some of my classmates consider more difficult than this type of problem. Any help would be very much appreciated! Thanks again!

 

[mmm4444bot]

g(x) = x/(x+1)

\(\displaystyle \text{Domain:}\) All real numbers except x=-1,

Horizontal Asymptote \(\displaystyle : y\)=1,

Vertical Asymptote \(\displaystyle : x\)=-1

I added some stuff above in LaTex, to clarify.

g'(x)= 1/(x+1)^2 -->x+1=0 --> x=-1, Partition number= -1

Using a sign chart: g(x) 0 = g'(x) 1, g(x) -2=g'(x) 1 (-infinity, -1)=increasing and (-1, infinity)=decreasing.

I don't get the "chart" notation, but I'm thinking that you picked two test values, one on each side of the vertical asymptote, x = -2 and x = 0.

If you're trying to say that you calculated g`(-2) and got 1, then don't write -2=g'(x) 1.

When x is -2, we write g`(-2) = 1.

Not sure how you concluded that g(x) is decreasing within (-1, oo). The value of g`(x) is positive in this interval, as the test value g`(0) = +1 shows.

We have:

g`(-2) = 1 and g`(0) = 1

These are both positive, so g`(x) is always a positive number.

The function g(x) increases over its entire domain because all values of its derivative are positive.

Again, I'm not certain what you did on the remaining three questions, either, but your conclusions look correct for those. :cool:

 

[irishpump]

Thank you very much! When I went back and saw what I did, I used my inputs that should have gone into my g(x) equation, but I instead put into the derivative. I'm not exactly sure how I did that, but I know my paper was a complete mess lol Thank you again!

I added some stuff above in LaTex, to clarify.




I don't get the "chart" notation, but I'm thinking that you picked two test values, one on each side of the vertical asymptote, x = -2 and x = 0.

If you're trying to say that you calculated g`(-2) and got 1, then don't write -2=g'(x) 1.

When x is -2, we write g`(-2) = 1.

Not sure how you concluded that g(x) is decreasing within (-1, oo). The value of g`(x) is positive in this interval, as the test value g`(0) = +1 shows.

We have:

g`(-2) = 1 and g`(0) = 1

These are both positive, so g`(x) is always a positive number.

The function g(x) increases over its entire domain because all values of its derivative are positive.

Again, I'm not certain what you did on the remaining three questions, either, but your conclusions look correct for those. :cool: