已知函数f(x)=ax^3+cx(a>0)在X1,X2处分别取得极值f(x1),且x1-x2的绝对值为2,f(x1)-f(x2)=x2-x1(1)求f(x)的解析式(2)求函数f(x)的单调区间与极值
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![已知函数f(x)=ax^3+cx(a>0)在X1,X2处分别取得极值f(x1),且x1-x2的绝对值为2,f(x1)-f(x2)=x2-x1(1)求f(x)的解析式(2)求函数f(x)的单调区间与极值](/uploads/image/z/6875073-9-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dax%5E3%2Bcx%28a%3E0%29%E5%9C%A8X1%2CX2%E5%A4%84%E5%88%86%E5%88%AB%E5%8F%96%E5%BE%97%E6%9E%81%E5%80%BCf%28x1%29%2C%E4%B8%94x1-x2%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%80%BC%E4%B8%BA2%2Cf%28x1%29-f%28x2%29%3Dx2-x1%281%29%E6%B1%82f%28x%29%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F%EF%BC%882%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B4%E4%B8%8E%E6%9E%81%E5%80%BC)
已知函数f(x)=ax^3+cx(a>0)在X1,X2处分别取得极值f(x1),且x1-x2的绝对值为2,f(x1)-f(x2)=x2-x1(1)求f(x)的解析式(2)求函数f(x)的单调区间与极值
已知函数f(x)=ax^3+cx(a>0)在X1,X2处分别取得极值f(x1),且x1-x2的绝对值为2,f(x1)-f(x2)=x2-x1
(1)求f(x)的解析式(2)求函数f(x)的单调区间与极值
已知函数f(x)=ax^3+cx(a>0)在X1,X2处分别取得极值f(x1),且x1-x2的绝对值为2,f(x1)-f(x2)=x2-x1(1)求f(x)的解析式(2)求函数f(x)的单调区间与极值
(1)函数f(x)=ax^3+cx(a>0)在X1,X2处分别取得极值
则得,f(x1)'=0,f(x2)'=0
f(x)'=3ax^2+c
则得3a(x1)^2+c=0,3a(x2)^2+c=o
两式相减得,3a(x1^2-x2^2)=0,3a(x1-x2)(x1+x2)=0
x1-x2的绝对值为2,a>0,则得x1+x2=0,又|x1-x2|=2,解得x1=1,x2=-1或x1=-1,x2=1
则得3a+c=0……①
又f(x1)-f(x2)=x2-x1
得a(x1)^3+c-a(x2)^3-c=x2-x1
化简得a(x1-x2){(x1)^2+x1x2+(x2)^2}+c(x1-x2)=x2-x1
为(x1-x2){a(x1)^2+ax1x2+a(x2)^2+c+1}=0
x1-x2≠0则得a(x1)^2+ax1x2+a(x2)^2+c+1=0
x1=1,x2=-1或x1=-1,x2=1
则得a+c+1=0……②
①-②得,2a=1,a=1/2,则得c=-3/2
则得f(x)=(1/2)x^3-(3/2)x
(2)f(x)'=(3/2)x^2-3/2
当f(x)=0,解得x=±1
则得,在(-∞,-1)和(1,+∞)为单调增区间,在(-1,1)为单调减区间
在x=1处取得极小值,极小值为-1
在x=-1处取的极大值,极大值为1