You can put this solution on YOUR website! What is the maxima and the minima to the nearest tenth of this equation,
-x^4+3x^2+3;-5<= x <=5?
DO YOU KNOW CALCULUS..PLEASE INFORM..THEN IT CAN BE EASILY DONE ...IF NOT .TRY THIS WAY...
Y= -{(X^2)^2-2*(3/2)*(X^2)+(3/2)^2}+(3/2)^2
=(9/4)-{X^2-1.5}^2
SINCE (X^2-1.5)^2 IS ALWAYS POSITIVE ITS MINIMUM VALUE IS ZERO.
S0 Y WILL BE MAXIMUM WHEN X^2=1.5..X=+ AND - SQUARE ROOT OF 1.5 WHICH IS IN THE GIVEN RANGE
MAXIMUM VALUE OF Y IS 9/4.