高一数学 大神来帮忙!!!第6.7题
展开全部
(6)
y= (2/5)^(-x^2-4x)
y'= (-2x-4) ln(2/5). (2/5)^(-x^2-4x)
y'=0
-2x-4 =0
x=-2
y'|x=-2+ <0
y'|x=-2- >0
x=-2 (min)
increasing interval = [-2,+∞)
decreasing interval =(-∞, -2]
(7)
(1)
f(x)= (a^x -1)/(a^x +1)
domain = R
f(x) =(a^x -1)/(a^x +1)
= 1 - 2/(a^x+1)
range = (-∞, 1)
(2)
f(-x) = (a^(-x) -1)/(a^(-x) +1)
= ( -1- a^x) /(1+a^x)
=-f(x)
f is odd function
(3)
f'(x) = (lna).a^x/(a^x+1)^2
case 1: 0<a<1
f'(x) = (lna).a^x/(a^x+1)^2 <0
decreasing interval =R
case 2: a>1
f'(x) = (lna).a^x/(a^x+1)^2 >0
increasing interval =R
y= (2/5)^(-x^2-4x)
y'= (-2x-4) ln(2/5). (2/5)^(-x^2-4x)
y'=0
-2x-4 =0
x=-2
y'|x=-2+ <0
y'|x=-2- >0
x=-2 (min)
increasing interval = [-2,+∞)
decreasing interval =(-∞, -2]
(7)
(1)
f(x)= (a^x -1)/(a^x +1)
domain = R
f(x) =(a^x -1)/(a^x +1)
= 1 - 2/(a^x+1)
range = (-∞, 1)
(2)
f(-x) = (a^(-x) -1)/(a^(-x) +1)
= ( -1- a^x) /(1+a^x)
=-f(x)
f is odd function
(3)
f'(x) = (lna).a^x/(a^x+1)^2
case 1: 0<a<1
f'(x) = (lna).a^x/(a^x+1)^2 <0
decreasing interval =R
case 2: a>1
f'(x) = (lna).a^x/(a^x+1)^2 >0
increasing interval =R
更多追问追答
追问
y'|x=-2+ 0
x=-2 (min)
这一步还是没懂 可以解释一下吗
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询