find the GLOBAL MAXIMUM AND...

tze

2011-07-30 2:55 am

find the GLOBAL MAXIMUM AND the global minimum of the function
y=(x^2-x+1)/(1+x+x^2)
for x bigger /equal than -2 and smaller or equal than 2

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回答 (1)

用戶2053

2011-07-30 4:48 am

✔ 最佳答案

y= (x² - x + 1) / (1 + x + x²)= (x² + x + 1 - 2x) / (1 + x + x²)= 1 - 2x / (1 + x + x²)= 1 - 2 / ( (x² + x + 1)/x )Let Y = (x² + x + 1)/x dY / dx = ( x(2x + 1) - (x² + x + 1) ) / x² = (x² - 1) / x²
and
d² Y / dx² = 2 / x³Let dY / dx = 0(x² - 1) / x² = 0x = 1 or x = - 1When x = 1 , d² Y / dx² > 0 ,
and hence Y(max.) = (1² + 1 + 1) / 1 = 3When x = - 1 , d² Y / dx² < 0 ,
and hence Y(min.) = ((-1)² - 1 + 1) / (- 1) = - 1Thus , y(min.) = 1 - 2 / Y(max.) = 1 - 2 / 3 = 1/3 (at x = 1 , - 2 ≤ x ≤ 2)y(max.) = 1 - 2 / Y(min.) = 1 - 2 / (-1) = 3 (at x = - 1 , - 2 ≤ x ≤ 2)

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