Finding number of real roots of an equation Equation is-->
$$ x^{13} + x - 1/e^x - \sin(x) =0 $$
To find number of real roots of the equation.
Context--> I am solving previous years questions of IIT Jam Mathematical 
Statistics (MS entrance exam) .
My approach-->
I took $e^{-x}$ and $\sin(x)$ to other side of the quation and expanded them. The coefficients of $x$ and $x^{13}$ were zero on RHS and if I put $x=0$ or $x=1$ there was no way to equate both sides. So I thought the answer would be no real roots. After that I put that graph on Desmos. Then I saw this graph cuts x axis at one point. So thank you for reading and answering this. 
Also my go to approach for finding real roots of a polynomial is to put values of $x$ and check the sign changes. Can you suggest me a different approach. I know about Descartes rule but that doesn't give exact number of real roots.
Example equation:-
$3x^3 - 12x^2 + 11x - 31=0$
 A: Setting
$f(x) = x^{13} + x - \dfrac{1}{e^x} - \sin x = x^{13} + x - e^{-x} - \sin x, \tag 1$
it is easy to see that there exists $M > 0$ such that
$x \le - M \Longrightarrow f(x) < 0, \tag 2$
and
$x \ge M \Longrightarrow f(x) > 0, \tag 3$
since the term $x^{13}$ dominates the others for large $x > 0$, as does $e^{-x}$ for large $x < 0$; it then follows from the intermediate value theorem that 
$\exists x_0 \in (-M, M), f(x_0) = 0; \tag 4$
thus, $f(x)$ has at least one real zero.  Next, we have
$f'(x) = 13 x^{12} + 1 + e^{-x} - \cos x > 0, \; \forall x \in \Bbb R, \tag 5$
since 
$1 + e^{-x} - \cos x > 0, \; \forall x \in \Bbb R, \tag 6$
and
$13 x^{12} \ge 0, \; \forall x \in \Bbb R, \tag 7$
it follows that
$f'(x) > 0, \; \forall x \in \Bbb R; \tag 8$
therefore $x_0$ is the sole real zero of $f(x)$.
As for polynomials $p(x) \in \Bbb R[x]$, for $\deg p(x) \ge 5$ there is not a lot to be said in general, but there are a few useful rules of thumb, e.g. $p(x)$ has at least one real root if $\deg p(x)$ is odd.  Of course, when $\deg p(x) \le 4$ one has well-known procedures and formulas to find solutions from the coefficients of $p(x)$.  In the case of
$3x^3 - 12x^2 + 11x - 31 = 0, \tag 9$ 
we may affirm the existence of a root; but to get much further, we might resort to Cardano's method which, though arithmetic intensive, is guaranteed to yield the desired results.
A: $$y(x)=x^{13}+x-e^{-x}-\sin(x)$$
$y(x)$is continuous (Sum of continuous functions). 
$$y'(x)=13x^{12}+1+e^{-x}-\cos(x)$$
$13x^{12}+1+e^{-x}>1\quad$ and $\quad-\cos(x)\geq -1\quad$ thus $y'(x)>0\quad$ any $x$.
$y(x)$  is continuous and strictly increasing from $-\infty$ to $+\infty$. 
Thus there is a root $f(x_0)=0$ with Bolzano's theorem and this root is unique.
Numerical solving (Newton-Raphson) :  $x_0\simeq 0.907676...$
