
#1
Dec1704, 11:15 AM

P: 137

Hi!
For example y=x^33x=0 gives y'=3x^23 and setting y'=0 we get i and i as the solutions. What does this say about the existence of the max and min points for the function y?  Kamataat 



#2
Dec1704, 11:26 AM

P: 23

Hi!
3x^{2}3=0 has no solutions in real numbers So, y' is always negative (as 3 is) Hence, y is always decreasing (no min and max) 



#3
Dec1804, 04:31 AM

P: 137

ok, thanks
 Kamataat 



#4
Dec1804, 09:42 PM

P: 84

maximaminima problems
x^33x=k
x^33xk=0=> where b makes x only have two solutions..... x^3+3x3x^2+k=(x^2bx(b/2)^2)(x+c) x^3bx^2b^4/4x+x^2cbcxcb^4/4 k= cb^4/4 b+c=0 b^4/4b^2=3 b^4+4b^2=12 b^4+4b^212=0 (b^2h)(b^2a) (a+h)=4 ah=12 



#5
Sep1710, 03:16 AM

P: 1

hi!
how determine whether that point is the maximum or the minimum? 



#6
Sep1710, 09:10 AM

Mentor
P: 21,012

There's more to this, but your calculus text should have more information about the details. In the future, if you have a question, start a new thread rather than adding onto an old thread. This thread is six years old. 


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