Finding roots of 5 th degree taylor expansion of $e^x$ I need to find the roots [actually their count and sign] of :
$$
{\mathcal T}(e^x,5)=\frac{x^5}{5!}+\frac{x^4}{4!}+\frac{x^3}{3!}+\frac{x^2}{2!}+x+1$$


*

*It[Root] is clearly negative.

*It should have odd number of roots.

*Either One or Three or Five.

*It is easy to find the derivatives, but only the third derivative, a quadratic can be easily analyzed.


Any help?
Basics Only.

A similar way to @Daniel Fischer's answer (if not quite as clean) to show that the derivative is positive:
$$\begin{aligned}
\frac{x^4}{4!} + \frac{x^3}{3!} + \frac{x^2}{2!} + x + 1 &= \frac{1}{24}\big(x^4+4x^3+12x^2+24x+24\big)\\
&= \frac{1}{24}\big((x^4+4x^3+6x^2+4x+1) + (6x^2+20x+23)\big)\\
&> \frac{1}{24}\big((x^4+4x^3+6x^2+4x+1) + (6x^2+20x+\frac{50}{3})\big)\\
&= \frac{1}{24}\big((x+1)^4 + 6(x+\frac53)^2\big)\\
&> 0.
\end{aligned}$$
