已知函数f(x)=e^(x-k) -x,其中X∈R
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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)