Exercise 1269Calculus › Single-variable calculus › Differentiation › Mean Value TheoremL1Added 4 Nov 2026

An exponential inequality

#mean value theorem#inequality#exponential

Statement

Show that for all $x>0$ : $e^x>1+x$ .

Hint

Apply the MVT to $e^x$ on $[0,x]$ ; the derivative $e^c>1$ for $c>0$ .

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MVT. There is $c\in]0,x[$ with $e^x-1=e^c\cdot x$ .

Bound. Since $c>0$ , $e^c>1$ , so $e^x-1>x$ : $e^x>1+x$ .

Conclusion. The inequality holds for $x>0$ .