Calculus of a Single Variable, Chapter 2, 2.2, Section 2.2, Problem 57

So I will start by rewriting this equation in the form:
f(x)=x^4-2x^2+3
We know that if we have then we can plug in any value for x, and the result will be what the slope of the function is at that given point.
So if we find the derivative of this original function, we should be left with the following:
f'(x)=4x^3-4x
If the tangent line to this graph is a horizontal line, then the slope should be 0/x or just 0.
So we can set what we calculated for f'(x) equal to zero and we can solve for our x-value:
4x^3-4x=0
We should simplify this to get the following:
4x(x^2-1)=0
4x=0 and (x^2-1)=0
x=0 and x-1=0 and x+1=0
So x should equal -1, 0, and 1
So, this tells us that if our "x" value is zero, one or negative one, then the slope should be horizontal. Now that we know the x-values of the coordinates, lets plug these into the original equation to find the y-coordinates!:
f(-1)=(-1)^4 -2(-1)^2+3=1-2+3=2
f(0)=(0)^4-2(0)^2+3 = 0+0+3 = 3
f(1)=(1)^4 - 2(1)^2 + 3 = 1-2+3 = 2

Therefore, our coordinates for the points where the tangent line to the graph are horizontal are:
(-1,2), (0,3), and (1,2)