F(X)是定义在[-1,1]上的偶函数,F(X)与G(X)的图像关于X=1对称,且当X∈[2,3]时g(x)=a(x-2)-2(x-2)^3
1、求F(X)的解析式2、若F(X)在[0,1]上是增函数,求实数a的取值范围3、当a∈,(-6,6),问能否使F(X)的最大值为4?请说明理由...
1、求F(X)的解析式
2、若F(X)在[0,1]上是增函数,求实数a的取值范围
3、当a∈,(-6,6),问能否使F(X)的最大值为4?请说明理由 展开
2、若F(X)在[0,1]上是增函数,求实数a的取值范围
3、当a∈,(-6,6),问能否使F(X)的最大值为4?请说明理由 展开
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解:
1)因为F(x)和G(x)关于x=1对称,所以F(1-x)=G(1+x),即是F(2-x)=G(x)。当X∈[2,3]时,2-x∈[-1,0]满足F(x)的定义,故F(2-x)=G(x)=a(x-2)-2(x-2)^3,令2-x=t,所以F(t)=2t^3-at,所以F(x)=2x^3-ax(x∈[-1,0])。又F(X)是定义在[-1,1]上的偶函数,f(x)=f(-x)=-2x^3+ax(x∈[0,1])。
2)F(X)在[0,1]上是增函数,所以F(x)的导数大于零,即是-2*3*x^2+a>0恒成立,解出来为a>6x^2,又因为x∈[0,1],所以a∈[0,6]
3)当a∈(-6,0)时,F(x)=2x^3-ax为增函数,有最大值f(0)=0,又因为F(X)为偶函数F(x)的最大值为0所以F(X)的最大值为4不成立。
1)因为F(x)和G(x)关于x=1对称,所以F(1-x)=G(1+x),即是F(2-x)=G(x)。当X∈[2,3]时,2-x∈[-1,0]满足F(x)的定义,故F(2-x)=G(x)=a(x-2)-2(x-2)^3,令2-x=t,所以F(t)=2t^3-at,所以F(x)=2x^3-ax(x∈[-1,0])。又F(X)是定义在[-1,1]上的偶函数,f(x)=f(-x)=-2x^3+ax(x∈[0,1])。
2)F(X)在[0,1]上是增函数,所以F(x)的导数大于零,即是-2*3*x^2+a>0恒成立,解出来为a>6x^2,又因为x∈[0,1],所以a∈[0,6]
3)当a∈(-6,0)时,F(x)=2x^3-ax为增函数,有最大值f(0)=0,又因为F(X)为偶函数F(x)的最大值为0所以F(X)的最大值为4不成立。
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FX是偶,再根据F,G对称.F从0到1就是上面的解析式。你可以画出大概的曲线图,就一目了然了,有点象|cosx|图。我解了N到题,,却没有一分,真预闷
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