What are the coordinates of the points where the graphs of f(x)=x^3-x^2+x+1 and g(x)=x^3+x^2+x-1 intersect? Give your answer as a list of points separated by semicolons. (So "(1,-3); (2,3); (5,-7)" is a valid answer format.)
To find the coordinates of the points where the graphs of f(x) and g(x) intersect, we need to find the values of x for which f(x) and g(x) are equal. This can be done by setting the two equations equal to each other and solving for x.
Setting f(x) = g(x), we have:
x^3 - x^2 + x + 1 = x^3 + x^2 + x - 1
Simplifying the equation, we get:
- x^2 + 2 = 2x^2 - 2
Bringing the terms to one side of the equation, we have:
3x^2 - x^2 = 2 + 2
Combining like terms, we get:
2x^2 = 4
Dividing both sides of the equation by 2, we have:
x^2 = 2
Taking the square root of both sides, we get:
x = ±√2
Therefore, the two x-values where f(x) and g(x) intersect are √2 and -√2.
Now we can substitute these x-values back into either equation to find the corresponding y-values.
For x = √2:
f(√2) = (√2)^3 - (√2)^2 + √2 + 1 = 2√2 - 2 + √2 + 1 = 3√2 - 1
g(√2) = (√2)^3 + (√2)^2 + √2 - 1 = 2√2 + 2 + √2 - 1 = 3√2 + 1
So, the coordinate of the point where the graphs intersect for x = √2 is (√2, 3√2 - 1).
For x = -√2:
f(-√2) = (-√2)^3 - (-√2)^2 - √2 + 1 = -2√2 - 2 - √2 + 1 = -3√2 - 1
g(-√2) = (-√2)^3 + (-√2)^2 - √2 - 1 = -2√2 + 2 - √2 - 1 = -3√2 + 1
So, the coordinate of the point where the graphs intersect for x = -√2 is (-√2, -3√2 + 1).
Therefore, the coordinates of the points where the graphs of f(x) and g(x) intersect are (√2, 3√2 - 1) and (-√2, -3√2 + 1).