1. ## Algorithm Sketching

I have questions in regards to:

Use the algorithm for curve sketching to sketch:

a) (2x+1)/(x-1)
b) (2x)/(x^2-25)

A) When I am figuring out the intervals of increase and decrease, I am supposed to use the first derivative test..
f'(x)=-3/(x-1)^2
and then find its critical numbers
f'(x)=0 when /
f'(x)=undefined when x=1
and then use the critical numbers to find where it is negative or positive, decreasing or increasing.
x<1 = decreasing
x>1 = decreasing
so f is always decreasing.

However, when I find the concavities, I find that the CU and CD intervals are not related to the intervals of decreasing and increasing..
f''(x)=6/(x-1)^3
critical numbers f''(x)=0 when x = /
f''(x)=undefined when x=1
Then do the second derivative test to find the concavities
x<0 = negative, concave down
x>0= positive, concave up

so when I say "f is always decreasing" above in my first derivative test, that doesn't match the second derivative results with CU and CD... I have a CU in there, but if concave up is positive, does it mean it has to increase? How can it increase if f is always decreasing.. OR are these completely different things? please clarify this for me.

I am wondering this similar question (can a concave up can occur when f is ALWAYS decreasing? and can a concave down occur when f is ALWAYS increasing?) with regards to part B as well.

2. Originally Posted by skeske1234
I have questions in regards to:

Use the algorithm for curve sketching to sketch:

a) (2x+1)/(x-1)
b) (2x)/(x^2-25)

A) When I am figuring out the intervals of increase and decrease, I am supposed to use the first derivative test..
f'(x)=-3/(x-1)^2
and then find its critical numbers
f'(x)=0 when /
f'(x)=undefined when x=1
and then use the critical numbers to find where it is negative or positive, decreasing or increasing.
x<1 = decreasing
x>1 = decreasing
so f is always decreasing.

However, when I find the concavities, I find that the CU and CD intervals are not related to the intervals of decreasing and increasing..
f''(x)=6/(x-1)^3
critical numbers f''(x)=0 when x = /
f''(x)=undefined when x=1
Then do the second derivative test to find the concavities
x<0 = negative, concave down
x>0= positive, concave up

so when I say "f is always decreasing" above in my first derivative test, that doesn't match the second derivative results with CU and CD... I have a CU in there, but if concave up is positive, does it mean it has to increase? How can it increase if f is always decreasing.. OR are these completely different things? please clarify this for me.

I am wondering this similar question (can a concave up can occur when f is ALWAYS decreasing? and can a concave down occur when f is ALWAYS increasing?) with regards to part B as well.

Yes, of course. For a simpler example, $\displaystyle y= x^3$ is always increasing but is concave down for x< 0 and concave up for x> 0. Conversely, $\displaystyle y= -x^3$ is always decreasing but is concave up for x< 0 and concave down for x> 0. "Concavity" does not have anything to do with "increasing" or "decreasing".