A question about combining derivatives and inequalities
2018.05.04
It is known that
\(e^x-(a+2)x \geqslant b-2\ ) is constant for any \(x \in R\) , then \(\dfrac{b-5}{a+2}\ ) has a maximum value of _______.
Analysis:
remember $f(x)=e^x-(a+2)x $, then
\[f'(x)=e^x-(a+2)\]
Since \(e^x-(a+2)x \geqslant b-2\ ) is constant for any \(x \in R\) , then the function \(f(x)\) must have a lower bound, so \( a+2>0\) .
At this point, \(f(x)\) decreases monotonically on \((-\infty,\ln (a+2))\) , and on \((\ln (a+2),+\infty)\ ) monotonically increasing.
Because
\[ f(\ln (a+2))=a+2-(a+2) \ln (a+2) \]
so
\[ a+2-(a+2) \ln (a+2 ) \geqslant b-2 \]
So
\[ \dfrac{b-5}{a+2} \leqslant \dfrac{(a+2)-(a+2)\ln (a+2) -3}{ a+2}=1-\ln(a+2)-\dfrac{3}{a+2} \]
Note \(g(x)=1-\ln x -\dfrac{3}{x}\ ) ,
Because
\[ g'(x)=-\dfrac{1}{x}+\dfrac{3}{x^2}=\dfrac{3-x}{x^2} \]
so \(g(x )\) increases monotonically on \((0,3)\) and decreases monotonically on \((3,+\infty)\) , thus
\[ g(x) \leqslant g(3)=-\ln 3 . \]
So
\[ \dfrac{b-5}{a+2} \leqslant -\ln 3 \]
And, at this time \(a=1,b=5-3\ln 3\) .