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Math Help - Differentiable Functions

  1. #1
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    Differentiable Functions

    I'm having a bit of trouble with this question

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

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

    3)\ Deduce\ that\ p_{4}(x)=1+x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}>0 for\ all\ x.
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  2. #2
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    Quote Originally Posted by acevipa View Post
    I'm having a bit of trouble with this question

    1)\ Show\ that\ the\ polynomial p_{3}(x)=1+x+ \frac{x^2}{2!} + \frac{x^3}{3!} has\ at\ least\ one\ real\ root
    If x= -100, p(-100)= 1- 100+ \frac{10000}{2}+\frac{-1000000}{6} = -99+ 5000- 16666 2/3< 0. If x= 100, 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?

    2)\ Show\ that\ the\ polynomial p_{2}(x)=1+x+ \frac{x^2}{2!} has\ no\ real\ roots\ and\ deduce that\ p_{3}(x)\ has\ exactly\ one\ real\ root.
    Use the quadratic formula. Then note that 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 \frac{x^3}{3!} causes the change- and that can only happen once.

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