What are the coordinates of the points where f(x) =x^3-x^2+x+1
And g(x)=x^3+x^2+x-1 intersect?
you want g(x) = g(x) , so
x^3 + x^2 + x - 1 = x^3 - x^2 + x + 1
2x^2 = 2
x^2 = 1
x = ± 1
if x = 1 , g(1) = 1 - 1 + 1 + 1 = 2
if x = -1 , g(-1) = -1 - 1 - 1 + 1 = -2
so they intersect at (1,2) and (-1,-2)
They intersect at two points. (1, 2) and (-1,-2).
If the graphs of $f$ and $g$ intersect at $(x,y)$, then $y = f(x)=g(x)$, so
\[x^3+x^2+x-1=x^3-x^2+x+1.\]Moving all the terms to the left-hand side gives $2x^2-2=0$, so $x=\pm1$. Substituting into $g$ we find the points of intersection to be
\[
(-1,g(-1))=(-1,-2)
\text{~and~}
(1,g(1))=(1,2).
\]Therefore, our answer is $\boxed{(-1,-2), (1,2)}$.
I did but I couldn't find the point of intersection.
He means you want f(x) to equal g(x).
I.e. f(x) = g(x) [NOT g(x) = g(x)]
When you set two equations equal to each other you find the points of intersection.