Intermediate-Value Theorem - Find roots of an equation I've an homework question where i need to prove that the following equation contains at-least three roots $ {x^4 \over 10} = {x^4-100  \over x-1} $.
I was able to find three roots after redefining the equation as function: $ f(x) =  {x^4-100 \over x-1} - {x^4 \over 10} $ where the segments:


*

*$ [3.3, 3.4]$ because $ f(3.3) > 0 $ and $f(3.4) < 0 $

*$ [10.9, 11]$ because $ f(10.9) > 0$ and $f(11) < 0$

*$ [-3, -2]$ because  $f(-3) < 0$ and $f(-2) > 0$


I know the Intermediate-Value theorem state that for each mentioned segment there has to be a 0 (a root).
I would like to know if my answer is correct and if so how can i prove that the function is continuous at the mentioned segments. 
UPDATE:
The roots are indeed in the above mentioned segments but in the open segment like:


*

*$(3.3, 3.4)$

*$(10.9, 11)$

*$(-3, -2)$

 A: Hint function is continuous at mentioned points if there  exists a limit at that point and its value is equal to the value  of the function at that point 
A: Assuming $x\neq 1$, the equation you have to solve is $$x^5-11 x^4+1000=0$$ So, consider the function and derivatives $$f(x)=x^5-11 x^4+1000$$ $$f'(x)=5x^4-44 x^3$$ $$f''(x)=20x^3-132x^2$$ The first derivative cancels at $x=0$ and $x=\frac{44}5$. $$f(\frac{44}5)=-\frac{38104056}{3125}<0$$ and the second derivative test shows that this is a minimum; so you have two real roots.
If you deflate the quintic polynomial, you are let with a cubic which has at least one real root.
So, you have at least three real roots that you properly located.
A: As said in another answer: assuming $x\neq1$, you have: $$
x^5-11 x^4+1000=0
$$
Now Descartes rule of signs tell you that this has exactly one negative root and either zero or two positive roots. As the left hand side is continuous and is positive at $x=1$ and negative at $x=10$, by the intermediate value theorem it is zero at at least one point in $(1,10)$. So there is at least one positive root and hence exactly two positive roots. So $x^5-11 x^4+1000=0$ has exactly three real roots ...
As $x=1$ is not a root for the original problem we have proven rather more than was asked, we have proven that there are exactly three roots for the original problem.
