作业帮f(x)在【a,b】上连续,f(a)=f(b)=0,一阶导数乘积大于零,证f(x)在[a,b]内至少有一个零点
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![f(x)在【a,b】上连续,f(a)=f(b)=0,一阶导数乘积大于零,证f(x)在[a,b]内至少有一个零点](/uploads/image/z/2624451-51-1.jpg?t=f%28x%29%E5%9C%A8%E3%80%90a%2Cb%E3%80%91%E4%B8%8A%E8%BF%9E%E7%BB%AD%2Cf%28a%29%3Df%28b%29%3D0%2C%E4%B8%80%E9%98%B6%E5%AF%BC%E6%95%B0%E4%B9%98%E7%A7%AF%E5%A4%A7%E4%BA%8E%E9%9B%B6%2C%E8%AF%81f%28x%29%E5%9C%A8%5Ba%2Cb%5D%E5%86%85%E8%87%B3%E5%B0%91%E6%9C%89%E4%B8%80%E4%B8%AA%E9%9B%B6%E7%82%B9)
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f(x)在【a,b】上连续,f(a)=f(b)=0,一阶导数乘积大于零,证f(x)在[a,b]内至少有一个零点
f(x)在【a,b】上连续,f(a)=f(b)=0,一阶导数乘积大于零,证f(x)在[a,b]内至少有一个零点
f(x)在【a,b】上连续,f(a)=f(b)=0,一阶导数乘积大于零,证f(x)在[a,b]内至少有一个零点
f'(a)f'(b)>0,不妨设f'(a)>0,f'(b)>0
则：lim[x→a+] [f(x)-f(a)]/(x-a)>0
由极限的局部保号性,存在a的右邻域(a,a+δ),使得当x∈(a,a+δ)时,有[f(x)-f(a)]/(x-a)>0
由于x>a,因此f(x)>f(a),在此邻域内取x1,则f(x1)>f(a)=0
同理可证：存在b的左邻域(b-δ,b),使得当x∈(b-δ,b)时,有[f(x)-f(b)]/(x-b)>0
由于x
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