Find the absolute maximum and absolute minimum, if either exists, for y = f(x) = x^4-8x^2+16, (A) [-2,1], (B) [0,2]

y = x^4-8x^2+16 = (x^2-4)^2

To find the absolute maximum and absolute minimum in [ A(-2,1) , B(0,2) ]

Solution:

We differentiate the function  and equate the function to zero. The solutions are the critical points. We also consider the end points.And evaluate for values of y within the given interval.

y' = 4x^3 + 16x .

y' = 0 gives 4x^3-16x = 0. Or x (4x^2 -16) = 0. Or x = 0. Or 4x^2 = 16. Or x^2 = 4. Or x = 2 or x = -2 the critical points.

But at the end points of [ A(-2,1) , B(0,2) ],  x = -2 and x = 0. So x = 2 is out side the interval and therefore is not of any consideration as far as the absolute  minimum is restricted to the given interval only.

So we find y values at the end points and the critical points obtained above.

So , y(-2) = (x-4)^2 = (4-4)^2 = 0

y(0) = (0-4)^2= 16,

y = 16 at x= 0,(an end point) is the absolute maximum at an

y = 0 at x =-2 , (also another end point) is the absolute minimum.