高中數學 求f(x)=(x^2-2x-1)(x^2-2x+3)+4x^2-8x+9之最小值?

f(x)
= (x² - 2x - 1)(x² - 2x + 3) + 4x² - 8x + 9
= (x² - 2x + 1 - 2)(x² - 2x + 1 + 2) + (4x² - 8x + 4) + 5
= [(x² - 2x + 1) - 2] [(x² - 2x + 1) + 2] + 4(x² - 2x + 1) + 5
= [(x - 1)² - 2] [(x - 1)² + 2] + 4(x - 1)² + 5
= {[(x - 1)²]² - 4} + 4(x - 1)² + 5
= [(x - 1)²]² + 4(x - 1)² + 1

由於 (x - 1)² ≥ 0，故此 f(x) = [(x - 1)²]² + 4(x - 1)² + 1 ≥ 1

f(x) 之最小值 = 1 ..... (答案)

f(x)=(x^2 -2x-1)(x^2 -2x+3)+4x^2 -8x+9
f'(x)=(x^2 -2x-1)(2x-2)+(x^2 -2x+3)(2x-2)+8x-8
f'(x)=(2x-2)(x^2 -2x-1+x^2 -2x+3+4)
f'(x)=2(x-1)(2x^2 -4x+6)
f'(x)=4(x-1)(x^2 -2x+3)
f'(0)=4(-1)(3)
f'(0)=-12
f(12)=15126
最小值為15126