一道常规的求参
已知函数\(f(x)=e^x-1\)
\((1)\) 若\(g(x)=f(x)-ax\),讨论\(g(x)\)的单调性
\((2)\)当\(x>0\)时,都有\((x-k-1)f(x)+x+1>0\)成立,求整数\(k\)的最大值
解
\((1)\) \(g(x)=e^x-1-ax\)
\(g^{\prime}(x)=e^x-a\)
当\(a\leq 1时,\)\(g^{\prime}(x)\geq 0\)
当\(a>1\)时,\(g^{\prime}(\ln a)=0\)
则\(g(x)\)在\(\left(0,\ln a\right)\)上单调递减,\((\ln a,+\infty)\)上单调递增
\((2)\) \((x-k-1)(e^x-1)+x+1>0\)
即\((x-k-1)e^x+k+2>0\)
记\(g(x)=(x-k-1)e^x+k+2\)
则\(g^{\prime}(x)=e^x(x-k)\)
若\(k\leq 0\)则\(g^{\prime}(x)\geq 0\)
从而\(g(x)>g(0)=1>0\)恒成立.
当\(k>0\)时,不难得到\(g(x)\)在\((0,k)\)单调递减,\((k,+\infty)\)单调递增
从而\(g(x)>g(k)=-e^k+k+2>0\)
记\(h(k)=k+2-e^k,h^{\prime}(k)=1-e^k<0\)
从而\(h(k)\)单调递减,又因\(h(1)=3-e>0,h(2)=4-e^2<0\)
从而\(k=1\)为题目所求的最大整数.