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Let f(x)={(x+2)3,−3<x≤−1x2/3,−1<x<2 and g(x)=x∫−3f(t) dt,−3<x<2. Then the number of extreme points of g′(x) is . |
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Answer» Let f(x)={(x+2)3,−3<x≤−1x2/3,−1<x<2 and g(x)=x∫−3f(t) dt,−3<x<2. Then the number of extreme points of g′(x) is . |
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