Ask your own question, for FREE!
Mathematics 23 Online
OpenStudy (precal):

Calculus Question: If I take the derivative of a function, and plot those critical points, do I test a number from each interval into the y' function

OpenStudy (turingtest):

you mean to find increasing/decreasing intervals? yes

OpenStudy (precal):

yes, to find inc and dec

OpenStudy (anonymous):

yes.

OpenStudy (precal):

thanks guys

OpenStudy (anonymous):

well for your info, if you plug values into y'', you will know whether the graph trend is concave upwards or downwards. that will help you in sketching graph too.

OpenStudy (precal):

thanks

OpenStudy (anonymous):

\[ y=4 x^5-45 x^4+180 x^3-310 x^2+240 x\\ y'= 20 x^4-180 x^3+540 x^2-620 x+240=\\ y'=20 (x-4) (x-3) (x-1)^2 \] You should not consider x=1 at all since (x-1)^2 is greater or equal to zero. To study the sign of y' in this case you study the sign of (x-4)(x-3) only.

OpenStudy (precal):

why would we not include x=1, it is a critical number

OpenStudy (precal):

the whole purpose of sign analysis is that we can talk about the graph of the original function without actually graphing it, the days where technology did not exist

OpenStudy (anonymous):

x=1 is a critical point. it may be a max/min point. once you plug into y'' you will know the nature of it.

OpenStudy (anonymous):

At x=1, you cannot have a maximum or a minimum. The derivative has the same sign before 1 and after 1. So it either increasing thru 1 or decreasing thru 1. To have a max say, f has to be increasing before 1 and decreasing after 1.

OpenStudy (anonymous):

x=1 can be an inflection point. Actually it is.

OpenStudy (anonymous):

yea sorry it's not a max/min point. but if i'm nt mistaken it's an inflection point.

OpenStudy (anonymous):

yes the value of y'' changes from - to +.

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!