Question

Find the absolute minimum and absolute maximum values of f on the given interval. f(x) = (x2 − 1)3, [−1, 2]

Answer 1

Given :

$F(x)=3(x^2-1)$

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.

$\dfrac{d(3x^2-3)}{dx}=0\\\\6x=0\\x=0$

So , x=0 is critical point .

Now , coefficient of $x^2$ is positive .

Therefore , it is increasing function after  x=0 .

So , min value will be at , x=0.

$Min = (0^2-1)\times 3=-3$

And maximum value will be the maximum at the x=2 because it is increasing function .

$Max=(2^2-1)\times 3=9$

Therefore , max and min value is 9 and -3 respectively .

Hence , this is the required solution .