<s>[INST] Prove that $-\frac {x}{2}\leq f(x)-1\leq \frac{x^2}{6}-\frac{x}{2}$ where $f(x)= \frac{1-e^{-x}}{x}$ Let $$\begin{cases}
f(x)= \frac{1-e^{-x}}{x},  & \text{if $x > 0$} \\
f(0)=1
\end{cases}$$
I have to prove that $\forall x \in \Bbb{R_+}-\frac {x}{2}\leq f(x)-1\leq \frac{x^2}{6}-\frac{x}{2}$ so i can show that $f'_d(0)=-\frac{1}{2}$ 
I tried using the mean value inequality but it didn't get me anywhere.
 [/INST] The given inequality is equivalent to: for $x>0$,
$$x\left(1-\frac {x}{2}\right)\leq 1-e^{-x}\leq x\left(1+\frac{x^2}{6}-\frac{x}{2}\right),$$
that is
$$1-x\left(1+\frac{x^2}{6}-\frac{x}{2}\right)\leq e^{-x}\leq 1-x\left(1-\frac {x}{2}\right)$$
or
$$1-x+\frac {x^2}{2}-\frac{x^3}{6}\leq e^{-x}\leq 1-x+\frac {x^2}{2}.$$
Now, you may apply the Lagrange's Remainder of the Taylor's expansion of $e^{-x}$ centered at $x_0=0$:  $\exists t,s\in (0,x)$ such that
$$e^{-x}=1-x+\frac {x^2}{2}-\frac{t^3}{6}=
1-x+\frac {x^2}{2}-\frac{x^3}{6}+\frac{s^4}{24}.$$
Can you take it from here? 
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