We have to determine the extreme points of f(x) = x^4 - 2x^2 + 1.

At the extreme points of the graph representing f(x), the value of f'(x) = 0.

For the function given f'(x) = 4x^3 - 4x

Solving 4x^3 - 4x = 0

=> x^3 - x = 0

=> x(x^2 - 1) = 0

=> x = 0, x = 1 and x = -1

The extreme points of the function f(x) = x^4 - 2x^2 + 1 are at x = 0, x = 1 and x = -1.

f''(x) = 12x^2 - 4

f''(0) = -4 , as the value of f''(0) is negative we have a maximum at x = 0

f''(1) = f''(-1) = 12 - 4 = 8, as this is positive we have a minimum at x = 1 and x = -1.

**The function f(x) = x^4 - 2x^2 + 1 has a minimum value at x = 1 and x = -1 and a maximum at x = 0**

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