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]
(The entire section contains 12 words.)
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) ]
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.