# Prove $f(x) =\ \frac{x^5}{5!} +\frac{x^4}{4!} +\frac{x^3}{3!}+\frac{x^2}{2!} +x+1$ has only one root.

We have to prove that the equation $\displaystyle \frac{x^5}{5!} +\frac{x^4}{4!} +\frac{x^3}{3!}+\frac{x^2}{2!} +x+1=0$ have exactly one real root .

My sir told me it is just an application of derivative .

But I could not understand what he mean by that . Can anybody please explain me .

• Maybe he just meant find $f'(x)$ and equate it to 0? – Iti Shree Apr 29 '17 at 17:13
• Hint: a monotonic function can only have one root. – Yves Daoust Apr 29 '17 at 17:14
• @ItiShree what will happen from that – user123733 Apr 29 '17 at 17:29

## 2 Answers

With a little inspection you can note that you have a truncated exponential function, which is monotonically increasing, and as pointed out can only have one real root. However, letting $f(x) = \frac{x^5}{120} + \frac{x^4}{24} + \frac{x^3}{6} + \frac{x^2}{2} + x + 1$, and noting that $$f'(x) = \frac{x^4}{24} + \frac{x^3}{6} + \frac{x^2}{2} + x + 1$$ Setting $f'(x) = 0$ we see that we have no real solutions. So there are no critical numbers, which shows the slope of the function does not change signs (either always increasing or decreasing), in this case always increasing. It suffices to show that the function $f(x)$ takes on a negative value and a positive value, and concluding by the intermediate value theorem that there exists a point in between those two values where it crosses the $x$-axis. We have that

$f(-3) = \dfrac{-13}{20}$ and $f(0) = 1$, so your real root lies between $-3$ and $0$.

• You could have been done with just using IVT and Rolle's theorem, with discussing monotonicity, nevertheless +1. – A---B Apr 29 '17 at 19:02
• "Setting $f'(x) =0$ we see that we have no real solutions" Is that something one should see immediately? To be honest, I don't. Did you mean after using the degree-4-polynomial-formula perhaps? Or are there much easier ways to see it? – Shashi Feb 4 at 20:12

Let's name the sequence of Taylor polynomials for $e^x,$ so $$f_0(x) = 1,$$ $$f_1(x) = 1 + x,$$ $$f_2(x) = 1 + x + \frac{x^2}{2},$$ $$f_3(x) = 1 + x + \frac{x^2}{2}+ \frac{x^3}{6},$$ $$f_4(x) = 1 + x + \frac{x^2}{2}+ \frac{x^3}{6}+ \frac{x^4}{24},$$ $$f_5(x) = 1 + x + \frac{x^2}{2}+ \frac{x^3}{6}+ \frac{x^4}{24}+ \frac{x^5}{120},$$ $$f_n(x) = 1 + x + \frac{x^2}{2}+ \frac{x^3}{6}+ \frac{x^4}{24}+ \frac{x^5}{120} + \cdots + \frac{x^n}{n!}.$$

The simple result is that, for $n$ even, we always have $f_n$ positive, no real roots. For $n$ odd, exactly one real root. This is induction, with the important relationships $$f_n'(x) = f_{n-1}(x),$$ $$f_n(x) = f_n'(x) + \frac{x^n}{n!}.$$

Induction, beginning with even: $f_{2n}(x)$ has a single minimum at the point $t$ where $f_{2n-1}(x)$ has its root. This root is never zero. However, there $$f_{2n}(t) = f_{2n-1}(t) + \frac{t^{2n}}{(2n)!} = \frac{t^{2n}}{(2n)!} > 0$$ We also need to point out that this means that $f_{2n+1}$ is always increasing, and, being a polynomial of odd degree, has exactly one real root. The cases $2n, 2n+1$ between them complete the induction step.