Prove that the equation $$\ x^4 = 18 + \frac{1}{1-x} $$ has at least three roots.
How do I use the intermediate value theorem to prove this? Does it mean I just need to find three different values for which $\ f(x) = x^4 - 18 - \frac{1}{1-x} $ is positive and therefore it will have a root?
On
Using Descartes Rule of signs on the polynomial function
$$ f(x)=x^5-x^4-18x+19 $$
we see that there are two changes of sign, thus there are either two or zero positive roots. But $f(0)$ is positive and $f(2)$ is negative. So it must be the case that there are two positive roots.
$$ f(-x)=-x^5-x^4+18x+19 $$
has only one change of sign in its coefficients, therefore there is one and only one negative root of the polynomial $f(x)$.
Thus the equation has two positive and one negative solutions.
Let $$g(x)= x^5-x^4-18x+19$$ Since we are looking for solution of $g(x)=0$ and we have $g(-3)<0$, $g(0)>0$, $g(2)<0$ and $g(3)>0$ we are done.