设x小于0时,f(x)=〔a(1-cos x)〕/x^2 x大于等于0时f(x)=x^2+bx+1求a,b的值在正负无穷上均连续,处处可导
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![设x小于0时,f(x)=〔a(1-cos x)〕/x^2 x大于等于0时f(x)=x^2+bx+1求a,b的值在正负无穷上均连续,处处可导](/uploads/image/z/5195038-22-8.jpg?t=%E8%AE%BEx%E5%B0%8F%E4%BA%8E0%E6%97%B6%2Cf%28x%29%EF%BC%9D%E3%80%94a%281-cos+x%29%E3%80%95%2Fx%5E2+x%E5%A4%A7%E4%BA%8E%E7%AD%89%E4%BA%8E0%E6%97%B6f%28x%29%3Dx%5E2%2Bbx%2B1%E6%B1%82a%2Cb%E7%9A%84%E5%80%BC%E5%9C%A8%E6%AD%A3%E8%B4%9F%E6%97%A0%E7%A9%B7%E4%B8%8A%E5%9D%87%E8%BF%9E%E7%BB%AD%EF%BC%8C%E5%A4%84%E5%A4%84%E5%8F%AF%E5%AF%BC)
设x小于0时,f(x)=〔a(1-cos x)〕/x^2 x大于等于0时f(x)=x^2+bx+1求a,b的值在正负无穷上均连续,处处可导
设x小于0时,f(x)=〔a(1-cos x)〕/x^2 x大于等于0时
f(x)=x^2+bx+1
求a,b的值
在正负无穷上均连续,处处可导
设x小于0时,f(x)=〔a(1-cos x)〕/x^2 x大于等于0时f(x)=x^2+bx+1求a,b的值在正负无穷上均连续,处处可导
f(0)=1
因为cosx= [cos(x/2)]^2-[sin(x/2)]^2
所以x
函数在0点连续时: 左边极限 = 右边极限
当x→-0时, lim[a(1-cosx)]/x²
=lim[a×1/2×x²]/x²
=a/2
当x→+0时, lim x²+bx+1 =1
所以,a/2 = 1
a = 2
可导时...
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函数在0点连续时: 左边极限 = 右边极限
当x→-0时, lim[a(1-cosx)]/x²
=lim[a×1/2×x²]/x²
=a/2
当x→+0时, lim x²+bx+1 =1
所以,a/2 = 1
a = 2
可导时,左导=右导
当x<0时,f′(x) = [2x²sinx - 4x(1-cox)]/x^4
= 2sinx/x² - 4(1-cosx)/x³
f′(0)= lim(2sinx/x²)-lim4(1-cosx)/x³
= lim 2/x - lim 2/x =0
当x>0时。f′(x)= 2x + b
f′(0) =b
所以 b=0
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