Question 1209334
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f(x) = g(x)
x^3-x^2+x+1 = x^3+x^2+x-1
x^3-x^2+x+1 - (x^3+x^2+x-1) = 0
x^3-x^2+x+1 - x^3-x^2-x+1 = 0
(x^3-x^3) + (-x^2-x^2) + (x-x) + (1+1) = 0
-2x^2+2 = 0
-2x^2 = -2
x^2 = 1
x = sqrt(1) or x = -sqrt(1)
x = 1 or x = -1


If x = 1, then,
f(x) = x^3-x^2+x+1 
f(1) = 1^3-1^2+1+1 
f(1) = 1-1+1+1 
f(1) = 2
You should find that g(1) = 2. I'll let you handle the scratch work for this portion.
This shows that (1,2) is a point of intersection.


If x = -1, then,
f(x) = x^3-x^2+x+1 
f(-1) = (-1)^3-(-1)^2+(-1)+1 
f(-1) = -1-1-1+1 
f(-1) = -2
You should find that g(-1) = -2. 


The two points of intersection are <font color=red>(1,2) and (-1,-2)</font>


Confirmation using a Desmos graph
<a href="https://www.desmos.com/calculator/bzmkmv82p3">https://www.desmos.com/calculator/bzmkmv82p3</a>
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