Exponential Tangent Inequality

Definition

Exponential Tangent Inequality

For every $x\in \mathbb{R}$, the natural exponential function lies above its tangent line at $0$:

$$e^x\ge 1+x.$$

Equality holds if and only if $x=0$. Equivalently, replacing $x$ by $-x$ gives

$$1-x\le e^{-x}.$$

Proof

Proof

Let $f(x)=e^x$. Since

$$f^{\prime \prime }(x)=e^x>0,$$

$f$ is strictly convex. A differentiable convex function lies above each of its tangent lines:

$$f(x)\ge f(a)+f^{\prime }(a)(x-a).$$

Taking $a=0$ yields

$$e^x\ge e^0+e^0(x-0)=1+x.$$

Strictly convexity makes equality possible only at the tangency point $x=0$. Substituting $-x$ for $x$ yields $1-x\le e^{-x}$.