I want to prove that $ e^x \ge 1+x $ for all $ x \in R $ , using Mean Value Theorem it can be proved for $ x \gt 0 $, and equality holds for $ x = 0$, however I can't solve it for $ x \lt 0 $
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consider the function $$f(x)=e^{x}-1-x$$ – Dr. Sonnhard Graubner Aug 19 '15 at 17:44
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2$g\colon x\mapsto e^x$ is convex, so its graph lies above the tangents. – Daniel Fischer Aug 19 '15 at 17:47
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Also, for $x < 0$ we have by the mean value theorem $e^x - 1 = x\cdot e^\xi > x$ for some $\xi \in (x,0)$, whence $0 < e^\xi < 1$ and thus $x\cdot e^\xi > x$. – Daniel Fischer Aug 19 '15 at 17:49
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First we take note that $e^0=1+0$ satisfies our inequality. Then as Dr. Sonnhard pointed out let us consider the function $f(x)=e^x-x-1$. The derivative of this function would be:
$$f'(x)=e^x-1$$
If we consider $x \leq 0$, this implies $e^x \leq 1$. Which means $e^x-1 \leq 0$, thus $f(x)$ is decreasing for any negative $x$.
So now we are sure (because $f(0)=0$ and $f(x)$ is decreasing as $x$ increases) that $f(x<0)>0$. So now we have.
For $x<0$, $e^x-x-1>0$
With some algebra:
For $x<0$, $e^x > x+1$
Ahmed S. Attaalla
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