设f(x)在[a b]上有二阶连续导数 且满足方程f〞(x)+x2fˊ(x)－2f(x)＝0 证明

正确答案：证明(用反证法)若最大值M＞0设f(x_M)＝Mx_M∈(ab)则由费马定理得fˊ(x_M)＝0又f(x_M)为极大值则f〞(x_M)＜0另由题设得f〞(x_M)＝－x^{2Mfˊ(x_M)+2f(x_M)＝2f(x_M)＝2M＞0(与f〞(x_M)＜0矛盾)故最大值M≤0同理可证最小值也必为0所以f(x)在[ab上的最大值M和最小值m都必为零因为f(a)＝f(b)＝0则f(x)在[ab上恒为零
证明(用反证法)若最大值M＞0,设f(xM)＝M,xM∈(a,b)则由费马定理得fˊ(xM)＝0,又f(xM)为极大值则f〞(xM)＜0,另由题设得f〞(xM)＝－x2Mfˊ(xM)+2f(xM)＝2f(xM)＝2M＞0(与f〞(xM)＜0矛盾)故最大值M≤0同理可证最小值也必为0,所以f(x)在[a,b上的最大值M和最小值m都必为零因为f(a)＝f(b)＝0,则f(x)在[a,b上恒为零}