已知函数f(x)=e^(x-k) -x,其中X∈R
1.当k=0时,若g(x)=1/[f(x)+m]定义域为R,求m的取值范围.
k=0
f(x)=e^(x-0)-x=e^x-x
g(x)=1/(e^x-x+m)
令h(x)=e^x-x+m
h’(x)=e^x-1=0
e^x=1 x=0
h’’(x)=e^x>0
h(0)为最小值
h(0)=e^0-0+m=1+m>0
m>-1
e^x-x+m≠0
定义域为R
2.给出定理:若函数f(x)在区间[a,b]上连续,且f(a)*f(b)1时,函数f(x)在[k,2k]内是否存在零点.
f(x)=e^(x-k)-x
f(k)=e^(k-k)-k=1-k0
h(ln2)=e^ln2-2ln2=2-2ln2>0
h(k)=e^k-2k>0
f(k)f(2k)=(1-k)(e^k-2k)