Prove that $7I was asked by my teacher to prove that $7<e^2<8$ using only algebraic methods and knowing that $2<e<3$.
I don't know how to do this, where to start from, but I guess that I would need some kind of a function where: $f'(\xi )=\frac{f(b)-f(a)}{b-a}$ so as to exploit monotonicity of $f'$. Any hint?
 A: Use the (convergent) series $$e^x=1+ x + \frac{x^2}{2!}+\frac{x^3}{3!}+\cdots.$$
For $x=2$, all the terms are positive and to get the lower bound take the  partial sum using first 5 terms, $1+ 2+ \frac42 + \frac86 +\frac{16}{24}=7$. So $e^2>7$.
Now look at the omitted (infinite) tail, starting from $\frac{32}{120}$. It is term-wise bounded above by the  geometric series with ratio $\frac12$, and first term $\frac{32}{120}$. 
Using the formula $\frac{a}{1-r}$ for  the sum of an infinite geometric series with first term $a$ and common ratio $r$  we get  the tail sum to be bounded above by  $\frac{32}{120}\big/\frac12= \frac{64}{120}<1$.
So to the full series for $e^2$, split as the  sum of head and  tail is $e^2< 7+1$.
A: It's not possible.
Since you aren't allowed to use any special property of e, for all practical purposes you can replace it with the variable x in the inequality and treat the inequality like 2

This means that x=2.1 is a valid case. But $x^2=4.41$ which is less than 7.
So, given only this property about e, we cannot prove 7
