Derivative big question practice (2023 college entrance examination real questions)

Known function $f(x)=to(e^x+a)-x$
(1) Discussf ( x ) f(x)The monotonicity of$f$$($$x$$)$
(2) proves that when$a>When\; 0$ , verify:$f\left(x\right)>2\ln a+\frac{3}{2}$

Solution:
\quad(1) $f^{\prime }(x)=a{e}^x-1$

\qquad ①$a>0$时，$x=-\ln When\; a$ f ′ ( x )$f^{\prime }\left(x\right)=0$

$f\left(x\right)$ in$[-\ln a,+\infty )$ , monotonically increasing on$(-\infty ,-\ln a]$ on monotonically decreasing

\qquad ②$a\le 0$ ,$f\left(x\right)$ at$(-\infty ,+\infty )$ monotonically decreasing

\quad (2) From (1), $x=-\ln a$ when$f\left(x\right)$ takes the minimum value

\qquadThe title is proof $f(-\ln a)>2\ln a+\frac{3}{2}$

\qquad 即$1+a^2-\ln a>2\ln a+\frac{3}{2}$，$a^2-3\ln a-\frac{1}{2}>0$

\qquad令$g(a)=a^2-3\ln a-\frac{1}{2}$，则$g^{\prime }(a)=2a\_-\frac{3}{a}$

\qquad当$a=\frac{\sqrt{6}}{2}$When $g^{\prime }\left(a\right)=0$

$g\left(a\right)$ in$[\frac{\sqrt{6}}{2},+\infty )$ , monotonically increasing at$(-\infty ,\frac{\sqrt{6}}{2}]$ on monotonically decreasing

\qquad Let $g(a)\ge g(\frac{\sqrt{6}}{2})=1-\ln \frac{3\sqrt{6}}{4}>0$

\qquad即$a^2-3\ln a-\frac{1}{2}>0$

\qquadFrom this we get $f(x)>2\ln a+\frac{3}{2}$