Show $x^4-x^2-x+1 = \sin(x)$ has a solution in the interval $[0,1]$ and also $[1,2]$ $x^4-x^2-x+1 = \sin(x)$
We have this equation here that should have a solution in each of the intervals $[0,1]$ and $[1,2]$.
How would we find these two solutions? I've tried with bisection methods or Intermediate Value Theorem but feel like my understanding in those concepts aren't strong enough to get an answer.
Furthermore; How can we show that there are no real solutions of the equation with x outside the interval $[−1, 2]$? I feel like the previous solution would help with this.
Finally. We are asked to Apply Newton’s method with starting value $x_0 = 0.9$, and find a solution of the equation correct to 6 decimal places.
I'm aware Newtons Method takes the form $x_{n+1}=x_n - \frac{f(x_1)}{f'(x_n)}$ 
Does this mean we should solve our equation for 0, then sub in the relevant values for the Newton's Method Form at $f(0.9)$, then compute values until two successive values give identical answers up to the 6th decimal place?
Thank you all for reading in advance!
 A: $f(x)=x^4-x^2-x+1=x^2(x^2-1)-(x-1)=(x-1)(x^3+x^2-1)$
Then $f(0)=1$ and $\sin 0=0$. $f(1)=0$, and $\sin 1>0$
Then let, $g(x)=x^4-x^2-x+1-\sin x$. $g(0)=1>0$ and $g(1)=-\sin 1<0$ and $g(x)$ is a continuous function. By Intermediate value theorem there is a point $c\in(0,1)$, such that $g(c)=0$
Similar for the interval $[1,2]$. (Note that $g(1)<0$ and $g(2)=11-\sin 2>0$)
Visualization:
Plot for $f(x)$ and $\sin x$ in simultaneously:

Plot of $g(x)$:

A: Hint: note that you haven't been told to find the solutions, only to prove that they're there. The intermediate value theorem is just what you need for that.
As for part 2, you can easily see that for $x>2$, $x^4$ is way too large compared to the other terms (it is now than four times larger than any of them, and therefore larger than their sum), so there can be no solution. The same is true for $x<-2$.
However, for $x<-1$, some more sophisticated methods are needed. We can see that at $-1$, the left side is too large for the right side, and if you are familiar with derivatives, then it is clear that as we move to smaller values of $x$, the left side increases and the right side decreases, taking us further away from a solution.
A: Let's consider $f(x) = x^4-x^2-x+1-\sin(x)$ instead.
Looking at the $[0,1]$ interval we have $\sin{0}=0$ and $0^4-0^2-0+1=1$ at the beginning of the interval, so $f(0) = 1$, while $\sin{1}$ is some positive value and $1^4-1^2-1+1=0$. It follows from Darboux property of continuous functions (it states that if your function has positive value at one point, and a negative value at another point, it must have a value of zero somewhere in between) that there must exist a point in that interval where $f(x)=0$, because $f(0)=1$ and $f(1)<0$. As for the interval $[1,2]$, it follows similarly, since $f(2)=2^4-2^2-2+1-\sin{2}=11-\sin{2}$, and since $\sin{x} \leq1$ for all $x$, $f(2)>0$. Thus we can see that there are solutions in $[0,1]$ and $[1,2]$.
