已知函数f(x)在R上满足f(x)=2f(2-x)-x2+8x-8,则曲线y=f(x)在点(1,f(1))处的切线方程是f(x)=2f(2-x)-x2+8x-8令x=1f(1)=2f(1)-1+8-8f(1)=1切点(1,1)对x求导f'(x)=2f'(2-x)*(2-x)'-2x+8f'(x)=-2f'(2-x)-2x+8令x=1f(1)=-2f(1)-2+8f(1)=2即
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![已知函数f(x)在R上满足f(x)=2f(2-x)-x2+8x-8,则曲线y=f(x)在点(1,f(1))处的切线方程是f(x)=2f(2-x)-x2+8x-8令x=1f(1)=2f(1)-1+8-8f(1)=1切点(1,1)对x求导f'(x)=2f'(2-x)*(2-x)'-2x+8f'(x)=-2f'(2-x)-2x+8令x=1f(1)=-2f(1)-2+8f(1)=2即](/uploads/image/z/12528559-55-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8R%E4%B8%8A%E6%BB%A1%E8%B6%B3f%28x%29%3D2f%282-x%29-x2%2B8x-8%2C%E5%88%99%E6%9B%B2%E7%BA%BFy%3Df%28x%29%E5%9C%A8%E7%82%B9%EF%BC%881%2Cf%281%29%EF%BC%89%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B%E6%98%AFf%28x%29%3D2f%282-x%29-x2%2B8x-8%E4%BB%A4x%3D1f%281%29%3D2f%281%29-1%2B8-8f%281%29%3D1%E5%88%87%E7%82%B9%281%2C1%29%E5%AF%B9x%E6%B1%82%E5%AF%BCf%27%28x%29%3D2f%27%282-x%29%2A%282-x%29%27-2x%2B8f%27%28x%29%3D-2f%27%282-x%29-2x%2B8%E4%BB%A4x%3D1f%281%29%3D-2f%281%29-2%2B8f%281%29%3D2%E5%8D%B3)
已知函数f(x)在R上满足f(x)=2f(2-x)-x2+8x-8,则曲线y=f(x)在点(1,f(1))处的切线方程是f(x)=2f(2-x)-x2+8x-8令x=1f(1)=2f(1)-1+8-8f(1)=1切点(1,1)对x求导f'(x)=2f'(2-x)*(2-x)'-2x+8f'(x)=-2f'(2-x)-2x+8令x=1f(1)=-2f(1)-2+8f(1)=2即
已知函数f(x)在R上满足f(x)=2f(2-x)-x2+8x-8,则曲线y=f(x)在点(1,f(1))处的切线方程是
f(x)=2f(2-x)-x2+8x-8
令x=1
f(1)=2f(1)-1+8-8
f(1)=1
切点(1,1)
对x求导
f'(x)=2f'(2-x)*(2-x)'-2x+8
f'(x)=-2f'(2-x)-2x+8
令x=1
f(1)=-2f(1)-2+8
f(1)=2
即切线斜率是2
所以切线2x-y-1=0
有一步不太懂
对x求导
f'(x)=2f'(2-x)*(2-x)'-2x+8
右边不是应该2f’(2-x)就好了么
已知函数f(x)在R上满足f(x)=2f(2-x)-x2+8x-8,则曲线y=f(x)在点(1,f(1))处的切线方程是f(x)=2f(2-x)-x2+8x-8令x=1f(1)=2f(1)-1+8-8f(1)=1切点(1,1)对x求导f'(x)=2f'(2-x)*(2-x)'-2x+8f'(x)=-2f'(2-x)-2x+8令x=1f(1)=-2f(1)-2+8f(1)=2即
f(2-x)
这是复合函数,要用链式法则
即u=2-x
y=f(u)
则u'=-1
y'=f'(u)×u'=-f'(2-x)
这个是复合函数求导啊!(f(u(x)))'=f'(u(x))*u'(x)