若函数f(x)=√(x²+1)-ax(a>0)在区间[1,+∞)上单调递增,求a的取值范围
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/29 21:54:30
![若函数f(x)=√(x²+1)-ax(a>0)在区间[1,+∞)上单调递增,求a的取值范围](/uploads/image/z/2114833-49-3.jpg?t=%E8%8B%A5%E5%87%BD%E6%95%B0f%28x%29%3D%E2%88%9A%28x%26%23178%3B%2B1%29-ax%28a%EF%BC%9E0%29%E5%9C%A8%E5%8C%BA%E9%97%B4%5B1%2C%2B%E2%88%9E%29%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%2C%E6%B1%82a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
若函数f(x)=√(x²+1)-ax(a>0)在区间[1,+∞)上单调递增,求a的取值范围
若函数f(x)=√(x²+1)-ax(a>0)在区间[1,+∞)上单调递增,求a的取值范围
若函数f(x)=√(x²+1)-ax(a>0)在区间[1,+∞)上单调递增,求a的取值范围
答:
f(x)=√(x^2+1)-ax(a>0)
显然,定义域为实数范围R
求导:
f'(x)=2x/[2√(x^2+1)]-a
=x/√(x^2+1)-a
f(x)在x>=1时是单调递增函数
所以:f'(x)=x/√(x^2+1)-a>=0在x>=1时恒成立
a<=x/√(x^2+1)
=√[x^2/(x^2+1)]
=√[1-1/(x^2+1)]
x>=1,x^2+1>=2
-1/2<=-1/(x^2+1)<0
1/2<=1-1/(x^2+1)<1
所以:a<=√(1/2)<=√[1-1/(x^2+1)]
所以:0
解:
f(x)=根号(x^2+1) - ax
f'(x)=1/2根号(x^2+1) - a=[1-2a根号(x^2+1)]/2根号(x^2+1)
令f'(x)>=0
则
1-2a根号(x^2+1)>=0
1>=2a根号(x^2+1)
1>=(4a^2)(x^2+1)
4a^2x^2+4a^2-1<=0
x^2<=1/4a^2 ...
全部展开
解:
f(x)=根号(x^2+1) - ax
f'(x)=1/2根号(x^2+1) - a=[1-2a根号(x^2+1)]/2根号(x^2+1)
令f'(x)>=0
则
1-2a根号(x^2+1)>=0
1>=2a根号(x^2+1)
1>=(4a^2)(x^2+1)
4a^2x^2+4a^2-1<=0
x^2<=1/4a^2 -1
-根号1/4a^2 -1<=x<=根号1/4a^2 -1时递增
所以根号1/4a^2 -1<=1
1/4a^2<=2
a∈(负无穷,-1/根号8]∪[1/根号8,正无穷)
收起