suppose f(x)=x^3/3-8xlnx+20x. show that there are no relative extreme points for this function.[hint:find the minimum value of the function's derivative]............help? not really sure what to do..

Question

dumbcow · Answer

im guessing you need to show that f'(x) cannot be 0, thus there are no local min/max values

f'(x) = x^2 - 8lnx+12

dumbcow · Answer

does that help?  do you see how that function is always greater than 0 given that domain is x>0

sirm3d · Answer

i have solved it. would you like to see my argument?

anonymous · Answer

ok i think i get it....sirm3d could u show me your argument anyways? please

sirm3d · Answer

my argument is that the graph of the derivative lies entirely in the first quadrant, although the graph is not needed.

sirm3d · Answer

i'll be back. gotta fetch a kid from school.

anonymous · Answer

ok that makes sense which means it would always be greater than zero

anonymous · Answer

ok than thanks!

sirm3d · Answer

what did you do in your solution?