Consider the function y = 3x5 – 25x3 + 60x + 1. Use the second derivative test to determine whether our function has a relative maximum or minimum at x = –1. Which of the following describes what you found?
(Points : 1)
x = –1 isn't a critical point, so I can't use the second derivative test, and there's neither a max nor a min there.
there, so there is a relative minimum at x = –1.
there, so there is a relative maximum at x = –1.
there, so there is neither a max nor a min there.
there, so the second derivative test fails, and I'd have to use a

Question

Consider the function y = 3x5 – 25x3 + 60x + 1. Use the second derivative test to determine whether our function has a relative maximum or minimum at x = –1. Which of the following describes what you found?
(Points : 1)
       x = –1 isn't a critical point, so I can't use the second derivative test, and there's neither a max nor a min there.
         there, so there is a relative minimum at x = –1.
         there, so there is a relative maximum at x = –1.
         there, so there is neither a max nor a min there.
         there, so the second derivative test fails, and I'd have to use a

campbell_st · Answer

why not just reply in the question you posted... rather than reposting 3 times...

anonymous · Answer

ITS A DIFFERENT QUESTION

campbell_st · Answer

wow...same curve... and given I gave you the factored form... who would have known...

anonymous · Answer

?

anonymous · Answer

well you have a slight error given the posting you just closed

f′(x)=15x4−75x2+60


which can be factored to 

f′(x)=15(x4−5x2+4)


which becomes

f′(x)(x2−1)(x2−4)


or 

f′(x)=(x−1)(x+1)(x−2)(x+2)


now you can play witht he 2nd derivative to check their nature

anonymous · Answer

THANKYOU

anonymous · Answer

MY BAD

anonymous · Answer

Consider the function y = 3x5 – 25x3 + 60x + 1. Use the first derivative test to decide whether this function has a maximum at x = 1.  Which of the following describes what you found?
(Points : 1)
       The derivative is positive to the left of x = 1 and negative to the right of x = 1, so the function has a relative minimum at x = 1.
       The derivative is positive to the left of x = 1 and negative to the right of x = 1, so the function has a relative maximum at x = 1.
       The derivative is positive to the left of x = 1 and positive to the right of x = 1, so the function has neither a relative maximum nor a minimum at x = 1.
       The derivative is negative to the left of x = 1 and positive to the right of x = 1, so the function has a relative minimum at x = 1.
       None of these apply.

campbell_st · Answer

here is a graph of your curve with with all the stationary points and points of inflexion shown.... oh... and infexion can also be  inflection... 

hope this helps with the new question

anonymous · Answer

ha! im working the same thing. my technique was using a number lick. pick a point aside of the critical point and see if the function is positive or negative, then you'll be able to tell if it's a max or min.