已知函数y=g(x)与f(x)=loga(x+1)(a>1)的图象关于原点对称(括号内为真数)(1)写出y=g(x)的解析式(2)若函数F(X)=F()x)+G(X)+M为奇函数,试确定实数M的值(3)当x∈[0,1)时,总有f(x)+g(x)≥n成立,求实数n的取值范围
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![已知函数y=g(x)与f(x)=loga(x+1)(a>1)的图象关于原点对称(括号内为真数)(1)写出y=g(x)的解析式(2)若函数F(X)=F()x)+G(X)+M为奇函数,试确定实数M的值(3)当x∈[0,1)时,总有f(x)+g(x)≥n成立,求实数n的取值范围](/uploads/image/z/2535314-50-4.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0y%3Dg%28x%29%E4%B8%8Ef%28x%29%3Dloga%28x%2B1%29%28a%3E1%29%E7%9A%84%E5%9B%BE%E8%B1%A1%E5%85%B3%E4%BA%8E%E5%8E%9F%E7%82%B9%E5%AF%B9%E7%A7%B0%28%E6%8B%AC%E5%8F%B7%E5%86%85%E4%B8%BA%E7%9C%9F%E6%95%B0%29%281%29%E5%86%99%E5%87%BAy%3Dg%28x%29%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F%282%29%E8%8B%A5%E5%87%BD%E6%95%B0F%28X%29%3DF%28%29x%29%2BG%28X%29%2BM%E4%B8%BA%E5%A5%87%E5%87%BD%E6%95%B0%2C%E8%AF%95%E7%A1%AE%E5%AE%9A%E5%AE%9E%E6%95%B0M%E7%9A%84%E5%80%BC%283%29%E5%BD%93x%E2%88%88%5B0%2C1%29%E6%97%B6%2C%E6%80%BB%E6%9C%89f%28x%29%2Bg%28x%29%E2%89%A5n%E6%88%90%E7%AB%8B%2C%E6%B1%82%E5%AE%9E%E6%95%B0n%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
已知函数y=g(x)与f(x)=loga(x+1)(a>1)的图象关于原点对称(括号内为真数)(1)写出y=g(x)的解析式(2)若函数F(X)=F()x)+G(X)+M为奇函数,试确定实数M的值(3)当x∈[0,1)时,总有f(x)+g(x)≥n成立,求实数n的取值范围
已知函数y=g(x)与f(x)=loga(x+1)(a>1)的图象关于原点对称(括号内为真数)
(1)写出y=g(x)的解析式
(2)若函数F(X)=F()x)+G(X)+M为奇函数,试确定实数M的值
(3)当x∈[0,1)时,总有f(x)+g(x)≥n成立,求实数n的取值范围
已知函数y=g(x)与f(x)=loga(x+1)(a>1)的图象关于原点对称(括号内为真数)(1)写出y=g(x)的解析式(2)若函数F(X)=F()x)+G(X)+M为奇函数,试确定实数M的值(3)当x∈[0,1)时,总有f(x)+g(x)≥n成立,求实数n的取值范围
(1)
图象关于原点对称,则g(x) + f(-x) = 0
g(x) = -f(-x) = -loga(1 - x)
(2)
F(x) = f(x) + g(x) = loga(x + 1) - loga(1 - x) + m = loga[(1+ x)/(1 - x)] + m
F(x)为奇函数,则F(-x) = -F(-x)
loga[(1- x)/(1 + x)] + m = -loga[(1+ x)/(1 - x)] - m
2m = -loga[(1+ x)/(1 - x)] - loga[(1- x)/(1 + x)]
= loga[(1 - x)/(1 + x)] - loga[(1- x)/(1 + x)]
= 0
m = 0
(3)
令G(x) = f(x) + g(x) = loga[(1+ x)/(1 - x)]
G'(x) =[(1- x)/(1 + x)]lna
a > 1,lna > 0
x∈[0,1)时:1-x > 0,1 + x > 0,G'(x) > 0
[0,1)上的最小值为G(0) = 0
n = 0