Analysis techinque in proving $e^x>1+x$ To prove that $e^x>1+x$, for all $x<0$. If I try to write $e^x$ as $1+x+\frac{1}{2}x^2+\frac{e^{\xi}}{3!}x^3$, for some $\xi\in (x,0)$. Then I get $e^x-(1+x)=\frac{1}{2}x^2+\frac{e^{\xi}}{3!}x^3$. How do I show that $\frac{1}{2}x^2+\frac{e^{\xi}}{3!}x^3$ is necessarily positive? I'd tried thinking for a long time, but can't find the way. Can't it be done by using Taylor to exactly order $3$??
 A: By using the shorter Taylor expansion, $e^x=1+x+\frac{e^{\xi}}{2!}x^2$ where $\xi$ is between $0$ and $x$, the proof is straightforward.
However, if you would like to consider the longer expansion $e^x=1+x+\frac{1}{2}x^2+\frac{e^{\xi}}{3!}x^3$, it suffices to show that for $x\in (-1,0)$,
$$\frac{1}{2}x^2+\frac{e^{\xi}}{3!}x^3=\frac{x^2}{6}\left(3+xe^{\xi}\right)>\frac{x^2}{6}\left(3+x\right)>0$$
which holds. Note that for $x\leq-1$, $e^x>0\geq 1+x$.
A: I don't think the approximation is the easiest way to go.
Rather, I would take a look at the function $f(x)=e^x - (1+x)$ and prove that the function has exactly one minimum which is acchieved at $x=0$.
A: If $x< 0,$ then $\int_x^0 e^t\,dt < \int_x^0 1\,dt,$ simply because $e^t < 1$ on $[x,0).$ Evaluate these integrals to see $1-e^{x} < -x,$ and the inequality follows.
A: For $u\in(0,1)$,
$$u-\ln u=1-\int_u^1\left(1-\frac1t\right)dt> 1$$
Then change $u=e^x$.
A: For $x<0$, $e^0=1+0$ and $(e^x)'<(1+x)'$. Starting from a common point, $e^x$ decreases slower than $1+x$.
A: Let $x \lt 0$ , $x$ real.
$f(x):= e^x$;
Mean Value Theorem:
$\dfrac{f(x) - f(0)}{x} =  $
$\dfrac{e^x -1 }{x} = f'(t)$ where , 
$t\in (x,0)$ for $x<0$.
Let $c: = e^t.$
Note :  $0 \lt c \lt 1.$
Hence: $e^x  = cx +1 > 1x +1.$
(Recall that $x\lt 0$.)
