# 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]

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

==> x^2= 4 ==> x = 2, -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.

The second part of the question giving two points (A) [-2, 1], and (B) [0, 2] is not clear. However we can find the point where value of the function f(x) = x^4 - 8x^2 + 16 as follows.

Derivative of f(x) = 4x^3 - 16x

= 4x(x^2 - 4) = 4x(x +2)(x - 2)

Maximum an minimum occur at point where the value of derivative is 0

Therefore:

4x(x + 2)(x - 2) = 0

Thus minimum and maximum values occur at values where:

x = 0, 2 or -2

Substituting these values of x in f(x) we get:

f(0) = 0^4 - 8*0^2 + 16 = 16

f(2) = 2^4 - 8*2^2 + 16 = 16 - 32 + 16 = 0

f(-2) = (-2)^4 - 8(-2)^2 + 16 = 16 - 32 + 16 = 0

Therefor:

Maximum value for f(x) is 16 which occurs at point (0, 16).

Minimum value for f(x) is 0 which occurs at two points (2, 0) and (-2, 0)