1\ x=2011 f(x)取到最值
由韋達定理,-b/a=4022 c/a=4044120
c=f(0)=8088240
所以a=2 b=-8044
最小值fmin=f(2011)=2(2011-2012)(2011-2010)=-2
2\ 將x=-1帶入,原式=2011-2009-2005-2003+2001=-2005
所以原式加上2005后,就含有因式x+1
又因為原式除以x+1的結(jié)果必然是一個常數(shù),所以這個常數(shù)是-2005,這就是余數(shù)
2道英文的數(shù)學題,
2道英文的數(shù)學題,
1)If f (x) = ax^2 + bx + c; f (2012) = 0; f (2010) = 0 and f (0) = 8,088,240
then for what value of x is f a maximum?What is the maximum value of f
2)What is the remainder when2011x^2010+ 2009x^1005+ 2005x^201+ 2003x^67
+ 2001 is divided by x + (Don’t use synthetic division!)
1)If f (x) = ax^2 + bx + c; f (2012) = 0; f (2010) = 0 and f (0) = 8,088,240
then for what value of x is f a maximum?What is the maximum value of f
2)What is the remainder when2011x^2010+ 2009x^1005+ 2005x^201+ 2003x^67
+ 2001 is divided by x + (Don’t use synthetic division!)
數(shù)學人氣：857 ℃時間：2020-06-10 21:30:27
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