设f(x)在[0,a]上连续,在(0,a)内可导,且f(0)=0,f(x)的导数单调增,证当0
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设f(x)在[0,a]上连续,在(0,a)内可导,且f(0)=0,f(x)的导数单调增,证当0
令g(x)=f(x)/x,x∈[0,a] g'(x)=[xf'(x)-f(x)]/x^2 另H(x)=xf'(x)-f(x) H'(x)=f'(x)+xf''(x)-f'(x)=xf''(x) ∵f(x)的导数单调递增 ∴f''(x)≥0 显然x>0 所以H'(x)≥0 ∴H(x)为在(0,a)单调递增 ∴H(x)≥H(0)=0-f(0)=0 ∴g'(x)≥0 ∴g(x)在(0,a)上单调递增 ∴当0
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