已知X=2时,多项式ax五次方+bx四次方+cx三次方+dx二次方+ex+f和bx四次方+dx二四方+f的字分别已知X=2时,多项式ax五次方+bx四次方+cx三次方+dx二次方+ex+f和bx四次方+dx二四方+f的字分别是4和3,则X=-2时,求多项式ax五次方+bx四次方+cx三次方+dx二次方+ex+f的值.
科比XKlx01
x=2时ax^5+bx^4+cx^3+dx^2+ex+f=4bx^4+dx^2+f=3两式相减：ax^5+cx^3+ex=1因此x=-2时ax^5+cx^3+ex=-1bx^4+dx^2+f=3ax^5+bx^4+cx^3+dx^2+ex+f=-1+3=2
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扫描下载二维码已知函数f(x)=x的四次方-4x的三次方+ax的平方-1在区间[0,1]上单调递增,在区间[1,2]上单调递减.(1) 若点A(X0 ,f（X0）)在函数f(x)的图象上,求证：点A关于直线X=1的对称点B也在函数f(x)的图像上；(2)是否存在实数b使得函数g(x)=x^4+bx^2+1(其中b小于4）的图像与f(x)的图象恰有3个交点,若有,求出实数b的范围,若不存在请说明理由.
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扫描下载二维码求函数y=x的四次方-4x的三次方+5的数值
这个题必须规定范围,若X取值范围（a,b)将函数进行两次求导,12X平方-24X 即12X（X-2) 求零点为0和2则原函数的导数4X三次方-12X平方在负无穷到0和2到正无穷为单增,而0到2为单减分别代入a ,b ,2.看看哪个小,且是否小于零若小于零则该点为函数的减区间点,代入原函数得出最小值再代入最大值点即可求最大值
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您给出一个函数，如果没有其他明确条件
如何能求出值呢？ 所以 您需要给定一个明确的 X取值或者 其他条件，否则只能因式分解~
你要做什么，说清楚
y=x^4-4x³+5=(x³-5)(x+1)
扫描下载二维码已知函数f(x)=x三次方+ax² 其中a属于R,讨论函数f(x)的单调性 ,没分了,来人呀
一旁的凌舟
f'（x）＝3ax^2＋2x＋bg（x）＝f（x）＋f‘（x）=ax^3＋x^2＋bx+3ax^2＋2x＋b=ax^3＋(1+3a)x^2＋(b+2)x＋b奇函数 (1+3a)=0
a=-1/3,b=0
f（x）=-1/3x^3+x^2
g(x)=-1/3x^3+2xg'(x)=-x^2+2
-根号2&x&根号2,增区间（-根号2,根号2）g'(x)&0
x&根号2z或x&-根号2 ,减区间（-无穷,-根号2）,（根号2,+无穷）g'(x)=0 x=根号2（设负）g（1）=5/3,g(2)=4/3,g(根号2)=4根号2/3最大值为4根号2/3,最小值为4/3
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扫描下载二维码已知函数f（x）=ax4+bcosx-x，且f（-3）=7，则f（3）的值为______．
承诺bGKZCN
由f（x）=ax4+bcosx-x，得f（x）+x=ax4+bcosx，设g（x）=ax4+bcos&x，则g（x）=g（-x）．即g（x）=f（x）+x是偶函数．由f（-3）=g（-3）+3，得g（-3）=f（-3）-3=4，∴g（3）=g（-3）=4，∴f（3）=g（3）-3=4-3=1．故答案为：1
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由f（x）=ax4+bcosx-x，得f（x）+x=ax4+bcosx，则f（x）+x是偶函数，利用函数的奇偶性，建立方程组即可求解．
本题考点：
函数奇偶性的性质．
考点点评：
本题主要考查函数值的计算，利用条件构造一个偶函数，利用偶函数的性质是解决本题的关键，
f（-3）=a乘（-3）^4+bcos（-3）+3=3^4乘a+bcos3+3=7∴3^4乘a+bcos3=4f（3）=a乘3^4+bcos3-3=4-3=1
f(3)=a(3)^4+bcos(3)-3=a(-3)^4+bcos(-3)+3-6=f(-3)-6=7-6=1
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