Finding the number of roots of the equation

Question

The equation $x^{13}-e^{-x}+x-\sin{x}=0$ has

1. No real root
2. More than two real roots.
3. Exactly two real roots.
4. Exactly one real root.

I tried doing with the odd derivative and check whether the derivative change sign or remain same(positive or negative) so i find $f^{(15)}(x)=e^{-x}+cos(x)$ now i am not able to determine how this will change the sign. Is there any other way to do the same problem thanks

Answer 1

Let $f:\mathbb{R}\to\mathbb{R}$ given by $f(x)=x^{13}-e^{-x}+x-\sin(x)$.

Then, $$\lim_{x\to +\infty}f(x)=\lim_{x\to +\infty}x^{13}-e^{-x}+x-\sin(x)\ge\lim_{x\to +\infty}x^{13}-e^{-x}+x-1=\infty$$

In the same way, $\lim_{x\to -\infty}f(x)=-\infty$.

More over $f^\prime(x)=13x^{12}+e^{-x}+1-\cos(x)>0$. Thus, $f$ is strictly increasing. By the Intermediate Value Theorem, $f$ has only one root.