1. ## Differentiable Functions

I'm having a bit of trouble with this question

$\displaystyle 1)\ Show\ that\ the\ polynomial$ $\displaystyle p_{3}(x)=1+x+ \frac{x^2}{2!} + \frac{x^3}{3!}$ $\displaystyle has\ at\ least\ one\ real\ root$

$\displaystyle 2)\ Show\ that\ the\ polynomial$ $\displaystyle p_{2}(x)=1+x+ \frac{x^2}{2!}$ $\displaystyle has\ no\ real\ roots\ and\ deduce$ $\displaystyle that\ p_{3}(x)\ has\ exactly\ one\ real\ root.$

$\displaystyle 3)\ Deduce\ that\ p_{4}(x)=1+x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}>0$ $\displaystyle for\ all\ x.$

2. Originally Posted by acevipa
I'm having a bit of trouble with this question

$\displaystyle 1)\ Show\ that\ the\ polynomial$ $\displaystyle p_{3}(x)=1+x+ \frac{x^2}{2!} + \frac{x^3}{3!}$ $\displaystyle has\ at\ least\ one\ real\ root$
If x= -100, $\displaystyle p(-100)= 1- 100+ \frac{10000}{2}+\frac{-1000000}{6}$$\displaystyle = -99+ 5000- 16666 2/3< 0$. If x= 100, $\displaystyle p(100)= 1+ 100+ \frac{10000}{2}+ \frac{1000000}{6}= 101+ 5000+ 16666 2/3> 0$. Since p is a polynomial, what does that tell you?

$\displaystyle 2)\ Show\ that\ the\ polynomial$ $\displaystyle p_{2}(x)=1+x+ \frac{x^2}{2!}$ $\displaystyle has\ no\ real\ roots\ and\ deduce$ $\displaystyle that\ p_{3}(x)\ has\ exactly\ one\ real\ root.$
Use the quadratic formula. Then note that $\displaystyle 1+ x+ \frac{x^2}{2!}$ is always of one sign so [tex]1+ x+ \frac{x^2}{2!}+ \frac{x^3}{3!}[tex] must change sign only when $\displaystyle \frac{x^3}{3!}$ causes the change- and that can only happen once.

$\displaystyle 3)\ Deduce\ that\ p_{4}(x)=1+x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}>0$ $\displaystyle for\ all\ x.$