数学已知函数f(x)=ax-ln(1+x^2) (1),当a=4/5时,求函数f(x)在(0,正无
1个回答

(1)f(x) = 4/5 x - ln(1+x^2)

f'(x) = 4/5 - 2x/(1+x^2) = 2(2x² - 5x+2) / [5(1+x²)]

令 f'(x) = 0 得 x= 1/2 ,2

x (0,1/2) 1/2 (1/2,2) 2 (2,+∞)

f'(x) >0 0 0

f(x) 递增 极大值 递减 极小值 递增

极大值 f(1/2) = 2/5 - ln(5/4)

极小值 f(2) = 8/5 - ln5

(2)当a=1时,f(x) = x - ln(1+x²)

f'(x) = 1 - 2x/(1+x²) = (x-1)² /(1+x²) > 0

f(x)是增函数,

∴f(x) > f(0) = 0

即 ln(1+x²) < x

(3)

ln(1+x²) < x

令 x = 1/n²

ln(1+ 1/n^4) < 1/n² < 1/[n(n-1)] = 1/(n-1) - 1/n

ln(1+1/2^4) < 1 - 1/2

ln(1+1/3^3) < 1/2 - 1/3

……

ln(1+1/n^4) < 1/(n-1) - 1/n

各式相加

ln [ (1+1/2^4)(1+1/3^4)…(1+1/n^4) ] < 1 -1/n < 1

即 (1+1/2^4)(1+1/3^4)…(1+1/n^4) < e