Given :
[tex]F(x)=3(x^2-1)[/tex]
To Find :
the absolute minimum and absolute maximum values of f on the given interval
[-1,2] .
Solution :
Now , getting first order differential equation and equating its equal to zero.
[tex]\dfrac{d(3x^2-3)}{dx}=0\\\\6x=0\\x=0[/tex]
So , x=0 is critical point .
Now , coefficient of [tex]x^2[/tex] is positive .
Therefore , it is increasing function after x=0 .
So , min value will be at , x=0.
[tex]Min = (0^2-1)\times 3=-3[/tex]
And maximum value will be the maximum at the x=2 because it is increasing function .
[tex]Max=(2^2-1)\times 3=9[/tex]
Therefore , max and min value is 9 and -3 respectively .
Hence , this is the required solution .
