Prove this inequality holds without using integrals We have a twice differentiable function $f: \mathbb{R} \to \mathbb{R}$ such that $f(2) = 3$, $f'(2) = 1$ and $f"(x) = \dfrac{e^{-x}}{x^2+1}$. Prove that $\dfrac{7}{2} \leq f(\dfrac{5}{2}) \leq \dfrac{7}{2} + \dfrac{e^{-2}}{40}$. 
I tried doing this with integrals but I quickly found that it was outside of my ability to obtain the necessary integral, so I figured there must be some simpler way to do this, but I don't really know how. 
edit: I kind of get the question now, but the $40$ in the denominator still eludes me, why is it there? 
 A: note that $\frac1{e^x(1+x^2)}$ is decreasing on the interval $[2, 5/2],$ therefore $$\frac{1}{2e} \le  \frac1{e^x(1+x^2)} \le \frac 1{5e^2}.$$  by the taylor theorem 
$$ \begin{align} f(5/2) &= f(2) + 1/2f'(2) + \int_2^{5/2}(x-2)f''(x)\, dx \\
 &\le 3 + 1/2+\frac 18f''(c) \text{ for some }2 < c < 5/2\\
\end{align}$$
therefore we have 
$$ \frac 1{16e} < f(5/2) - 7/2 \le \frac1{40e^2}  $$.
A: It can easily be checked that, for $x\ge 2$, we have
$$ 0\le f''(x) \le \frac{e^{-2}}{5}. $$
By the Fundamental Theorem of Calculus, we have $f'(x) = f'(2) + \int\limits_{2}^{x}{f''(t)\text{ d}t} = 1 + \int\limits_{2}^{x}{f''(t)\text{ d}t}$, so applying the above bounds gives
$$1 + \int\limits_{2}^{x}{0\text{ d}t}\le f'(x)\le 1+\int\limits_{2}^{x}{\frac{e^{-2}}{5}\text{ d}t} \implies 1\le f'(x)\le 1 + \frac{e^{-2}}{5}(x-2) $$
for $x\ge 2$. Applying FTC again yields
$$ f(2) + \int\limits_{2}^{x}{1\text{ d}t} \le f(x) \le f(2) + \int\limits_{2}^{x}{1+\frac{e^{-2}}{5}(x-2)\text{ d}x} $$
or
$$3 + (x-2) \le f(x) \le 3 + (x-2) + \frac{e^{-2}}{10}(x-2)^2.$$
Plugging in $x = \frac{5}{2}$ gives the desired result.
