设y=f(x)在(-∞,+∞)上连续且单调递减,试证:函数F(x)=∫ {0,x}(x-2t)f(t)dt 在(-∞,+∞)单调递F(x)=∫[0,x] (x-2t)f(t)dt=x∫[0,x] f(t)dt-2∫[0,x] tf(t)dtF'(x)=∫[0,x] f(t)dt+xf(x)-2xf(x)=∫[0,x] f(t)dt-xf(x)F''(x)=f(x
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![设y=f(x)在(-∞,+∞)上连续且单调递减,试证:函数F(x)=∫ {0,x}(x-2t)f(t)dt 在(-∞,+∞)单调递F(x)=∫[0,x] (x-2t)f(t)dt=x∫[0,x] f(t)dt-2∫[0,x] tf(t)dtF'(x)=∫[0,x] f(t)dt+xf(x)-2xf(x)=∫[0,x] f(t)dt-xf(x)F''(x)=f(x](/uploads/image/z/5124065-41-5.jpg?t=%E8%AE%BEy%3Df%28x%29%E5%9C%A8%EF%BC%88-%E2%88%9E%2C%2B%E2%88%9E%EF%BC%89%E4%B8%8A%E8%BF%9E%E7%BB%AD%E4%B8%94%E5%8D%95%E8%B0%83%E9%80%92%E5%87%8F%2C%E8%AF%95%E8%AF%81%EF%BC%9A%E5%87%BD%E6%95%B0F%28x%29%3D%E2%88%AB+%7B0%2Cx%7D%EF%BC%88x-2t%29f%28t%29dt+%E5%9C%A8%EF%BC%88-%E2%88%9E%2C%2B%E2%88%9E%EF%BC%89%E5%8D%95%E8%B0%83%E9%80%92F%28x%29%3D%E2%88%AB%5B0%2Cx%5D+%28x-2t%29f%28t%29dt%3Dx%E2%88%AB%5B0%2Cx%5D+f%28t%29dt-2%E2%88%AB%5B0%2Cx%5D+tf%28t%29dtF%27%28x%29%3D%E2%88%AB%5B0%2Cx%5D+f%28t%29dt%2Bxf%28x%29-2xf%28x%29%3D%E2%88%AB%5B0%2Cx%5D+f%28t%29dt-xf%28x%29F%27%27%28x%29%3Df%28x)
设y=f(x)在(-∞,+∞)上连续且单调递减,试证:函数F(x)=∫ {0,x}(x-2t)f(t)dt 在(-∞,+∞)单调递F(x)=∫[0,x] (x-2t)f(t)dt=x∫[0,x] f(t)dt-2∫[0,x] tf(t)dtF'(x)=∫[0,x] f(t)dt+xf(x)-2xf(x)=∫[0,x] f(t)dt-xf(x)F''(x)=f(x
设y=f(x)在(-∞,+∞)上连续且单调递减,试证:函数F(x)=∫ {0,x}(x-2t)f(t)dt 在(-∞,+∞)单调递
F(x)=∫[0,x] (x-2t)f(t)dt=x∫[0,x] f(t)dt-2∫[0,x] tf(t)dt
F'(x)=∫[0,x] f(t)dt+xf(x)-2xf(x)=∫[0,x] f(t)dt-xf(x)
F''(x)=f(x)-f(x)-xf'(x)=-xf'(x) 由题意知f'(x)0时,F''(x)>0,x
设y=f(x)在(-∞,+∞)上连续且单调递减,试证:函数F(x)=∫ {0,x}(x-2t)f(t)dt 在(-∞,+∞)单调递F(x)=∫[0,x] (x-2t)f(t)dt=x∫[0,x] f(t)dt-2∫[0,x] tf(t)dtF'(x)=∫[0,x] f(t)dt+xf(x)-2xf(x)=∫[0,x] f(t)dt-xf(x)F''(x)=f(x
x>0时,F''(x)>0,x
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