I want to how nicely define the $f(x)$ for this type of question
to prove the inequality use the mean value theorem
$$e^x \ge 1+x ,\ x \in \mathbb{R}$$
How to choose $f(x)$ to show that inequality and do
$$f(b) - f(a) = f'(c)(b-a),\ \text{since}\ f:[a,b] \to \mathbb{R} $$ is continuous and differentiable
Bernoulli's Inequality says that for $n\ge1$ and $x\ge-n$ $$ \left(1+\frac xn\right)^n\ge1+x $$ Since $$ e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n $$ we get that for all $x\in\mathbb{R}$, $$ e^x\ge1+x $$
Let $f(x)=e^x-x$. Then for some $\xi$ between $0$ and $x$, the Mean Value Theorem says $$ \frac{(e^x-x)-1}{x-0}=e^\xi-1 $$ That is, $$ e^x-1-x=x(e^\xi-1) $$ Note that $x\gt0\iff e^\xi-1\gt0$; therefore, $$ e^x-1-x\ge0 $$
