Find the number of real solutions to the equation $ x^7=x+1$. The equation $x^7 = x+1 $ has:

a) no real solution
b)  no positive real solution
c) a real solution in the interval (0,2)
d) a real solution but not within (0,2)
Which is the correct answer and why ? How do i find the answer to such questions? 

To solve a question like this, let's use Descartes's rule of signs.
So our function is $f(x) = x^7 - x - 1 $
We then do the following:
$f(-x) = -x^7 + x - 1$
There is one change in sign for f(x), meaning there is one positive real root.
There are two changes for f(-x), meaning there are either 2 or 0 negative real roots.
Since there are 7 roots, by the fundamental theorem by algebra, we could have 1 positive root, 2 negative roots, and 4 complex roots.
Or we could have 1 positive root, 0 negative roots, and 6 complex roots.
We know that the answer is either c or d.
So, we can check values.
$f(0) = -1$, and $f(2) = 2^7 - 2 -1 = 125$
Since from $(0,2)$ we have $f(x)$ change from negative and positive. Therefore, by the IVT (Intermediate Value theorem), there is at least one zero in that open interval.
The answer is c.
