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.

Consider the derivative
$$\begin{aligned}
\frac{x^4}{4!} + \frac{x^3}{3!} + \frac{x^2}{2!} + x + 1 &= \frac{1}{4!}(x^4 + 4x^3 + 4x^2) + \frac{1}{12}x^2 + \left(\frac{x^2}{4} + x + 1\right)\\
&= \frac{x^2}{24}(x+2)^2 + \frac{x^2}{12} + \left(\frac{x}{2}+1\right)^2\\
&> 0.
\end{aligned}$$
So the fifth-order Taylor polynomial is strictly increasing, hence has exactly one real zero.
