Question

how to solve the critical value and minima and maxima in this equation: y= x3-3x. plzzz... i need your answer...

Answer 1

take the derivative:
y'=3x^2-3, set it equal to 0:
3x^2-3=0; x^2-1=0; x=+/-1

Answer 2

check which one's a max with 2nd derivative test.

Answer 3

y''=6x

Answer 4

y''(1)=6>0=>x=1 is a min

Answer 5

y''(-1)=-6<0=>x=-1 is a max

Answer 6

is there any graph for that?..

Answer 7

yes. taking into consideration the max at x=1 and min at x=-1, and your zeros of f at x=0 and sqrt3 and -sqrt3, you can make a sketch...