Question 755154
The function f(x)=x^3-x+1 has exactly one real zero. It is between 
(A) -2 and -1
(B) -1 and 0 
(C) 0 and 1
(D) 1 and 2 
(E) 2 and 3 
Can you please tell me how to get this answer? its on my final and i don't remember ever doing them..
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There is a rule I found called the Location Theorem which can be applied here.
This is how it works: plug in consecutive x-values like a and b.
If f(a) and f(b) have opposite signs, there is at least one real root between a and b.
..
(A) a=-2, b=-1
f(a)=f(-2)=-8+2+1=-5
f(b)=f(-1)=-1+1+1=1
f(a) and f(b) have opposite signs, so there is a real root between -2 and -1
..
(B) a=-1, b=0
f(a)=f(-1)=1
f(b)=f(0)= 1
f(a) and f(b) have same signs, so there isno real root between -1 and 0
..
(C) a=0, b=1
f(a)=f(0)=1
f(b)=f(1)=1
f(a) and f(b) have same signs, so there is no real roots between 0 and 1
..
You can try (D) and (E)  yourself.
If you have a graphing calculator, you will find a real zero at -1.32472.., confirming that A is the ans.