设f(x)=ax²+bx+ln(x+1) (a＞0) 1°当b=0时，讨论函数单调性

设f(x)=ax²+bx+ln(x+1) (a＞0) 1°当b=0时，讨论函数单调性

1.
f(-1) = a - b + 1 = 0, b = a+1
且函数f(x)的值域为[0,﹢无穷), 显然x = -1为该抛物线的对称轴,且抛物线开口向上, a >0
对称轴x = -1 = -b/(2a) = -(a+1)/(2a)
a = 1
f(x)=x²+2x+1
F(x) = x²+2x+1, x > 0
= -(x²+2x+1), x < 0
2.
g(x) = x²+2x+1 - kx = x²+(2 - k)x+1
g(x)的对称轴为x = -(2-k)/(2*1) = (k-2)/2
当x∈[-2,2]时, g(x)是单调函数, 对称轴x = (k-2)/2 ≥2 或 x = (k-2)/2 ≤ -2
即k ≥ 6或k ≤ -2
3
不影响结果,不妨设m > 0
mn < 0, n < 0
m + n > 0, m > |n|, m² > n²
f(x)为偶函数, b = 0, f(x) = ax² + 1
F(m) = f(m) = am²+ 1
F(n) = -f(n) = -an² - 1
F(m)+F(n) = a(m² - n²) > 0