Hi! For example y=-x^3-3x=0 gives y'=-3x^2-3 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
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)
-x^3-3x=k -x^3-3x-k=0=> where b makes x only have two solutions..... x^3+3x-3x^2+k=(x^2-bx-(b/2)^2)(x+c) x^3-bx^2-b^4/4x+x^2c-bcx-cb^4/4 k= -cb^4/4 -b+c=0 -b^4/4-b^2=-3 b^4+4b^2=12 b^4+4b^2-12=0 (b^2-h)(b^2-a) (a+h)=-4 ah=12
The second derivative test is helpful. At a critical number c for which f'(c) = 0, if f''(c) > 0, (c, f(c)) is a local minimum point; if f''(c) < 0, (c, f(c)) is a local maximum point. 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.