Ask your own question, for FREE!
Mathematics 9 Online
OpenStudy (anonymous):

relative extrema of x^3(x+1)^2

OpenStudy (anonymous):

first differentiate it to get f'. then equate f'=0 to get the critical points. differentiate again to get f". check which point that will yield f">0. then the x is the relative minima.

OpenStudy (anonymous):

for relative extrema, check which x u got yield f"<0

OpenStudy (anonymous):

product rule right?

OpenStudy (anonymous):

yes.

OpenStudy (anonymous):

for the (x+2)^2 part, do i do (x+1)(x+1) and derive the product of that?

OpenStudy (anonymous):

no. use this: if f(x)= (g(x))^n then f'(x) = n*g'(x)*(g(x))^(n-1)

OpenStudy (anonymous):

3x(x+1)^2 + (2x+2)(x^3)?

OpenStudy (anonymous):

Yes except you forgot the square on your first x

OpenStudy (anonymous):

3x^2*(x+1)^2 + (2x+2)(x^3). u forgot the square at front.

OpenStudy (anonymous):

how about 2x+3x^(2/3)?

OpenStudy (anonymous):

i end up getting 2x^(-1/3) +2

OpenStudy (anonymous):

thats correct

OpenStudy (anonymous):

from there i get 2/cube root x +2?

OpenStudy (anonymous):

yes.

OpenStudy (anonymous):

how do you do absolute value problems?

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!