Ask your own question, for FREE!
Mathematics 11 Online
OpenStudy (anonymous):

PLEASE Calculus Help? (a) find the critical numbers of f (if any), (b) find the open interval(s) on which the function is increasing or decreasing, (c) apply the First Derivative Test to identify all relative extrema, and (d) use a graphing utility to confirm your results . 2. f(x)= x/x+1

OpenStudy (anonymous):

I already know I have to start by finding the derivative which is f'(x) = 1/ (x+1)^2 right?

OpenStudy (anonymous):

and the roots would be x=-1

OpenStudy (anonymous):

?

OpenStudy (alekos):

that derivative doesnt look right

OpenStudy (anonymous):

I did it wrong'?

OpenStudy (anonymous):

can you help me?

OpenStudy (alekos):

for the derivative you can use the quotient rule f'(u/v) = (u'v -v'u)/v^2 where u=x and v=x+1

OpenStudy (alekos):

the critical numbers would be x=0 and x=-1 because f(0) = 0 and f(-1) is undefined

OpenStudy (anonymous):

ok

OpenStudy (alekos):

also the lim x->inf = 1 and lim x->-inf = 1

OpenStudy (anonymous):

so would the derivative be x(x+1) - x+1(x)/(x+1)^2 ?

OpenStudy (anonymous):

is that the answer for b?

OpenStudy (alekos):

should be x+1 not x=1

OpenStudy (alekos):

x' = 1 and (x+1)' = 1 do you follow?

OpenStudy (anonymous):

I don't understand... is that part b? or are you talking about the derivative?

OpenStudy (alekos):

starting again it should be [x'(x+1) - (x+1)'x]/(x+1)^2

OpenStudy (loser66):

for supporting only, @alekos http://www.wolframalpha.com/input/?i=%28x%2F%28x%2B1%29%29%27

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

... I'm confused now.

OpenStudy (loser66):

your derivative is perfectly OK. f'(x) = \(\dfrac{1}{(1+x)^2}\)

OpenStudy (loser66):

and it never =0, it's undefined when denominator =0, it means x =-1, so far so good?

OpenStudy (anonymous):

yes.

OpenStudy (alekos):

my apologies you are right it comes to 1/(x+1)^2

OpenStudy (loser66):

ok, go ahead @alekos

OpenStudy (anonymous):

and the critical numbers are correct yes?

OpenStudy (alekos):

yes they are

OpenStudy (loser66):

it's another case to study, friend. 1 > 0 always, OK? now denominator, except x = -1, x +1 =0 for all values of x, (x+1)^2 >0 got this part?

OpenStudy (loser66):

for example, if x =-2, x +1 = -1, then (x+1)^2 = (-1)^2 =1 >0 if x = 5 , x +1 =6 then (x+1)^2 = 6^2 =36>0

OpenStudy (anonymous):

right. I got x=-1 and alekos showed me x=0. I understand that

OpenStudy (anonymous):

yes I get that

OpenStudy (loser66):

ok, now make the table |dw:1386949138165:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!