y^4 = y^2 - x^2
4y^3 y' = 2y y' - 2x
y' = -2x/(4y^3-2y)
Now, assuming that y ≠ 0 or ±1/√2,
y'=0 when x=0
so, y^4 = y^2, and y = ±1, so
(0,1) and (0,-1)
extra credit: what happens at y = 0 or ±1/√2?
determine the point(s)at which the graph of y^4 = y^2 -x^2 has a horizontal tangent.
5 answers
how did you know it was + or - l? I got what x equals but I was confused on how to find the y values?
if x=0,
y^4 = y^2
y^2(y^2-1) = 0
y = 0 or ±1
but y=0 is not a possibility.
y^4 = y^2
y^2(y^2-1) = 0
y = 0 or ±1
but y=0 is not a possibility.
where did you get y^2 -1 from? Did you subtract y^2 from the right side? if so where did the -1 come from?
Hello? This is calculus. Have you forgotten you algebra I?
y^4 = y^2
y^4 - y^2 = 0
y^2(y^2-1) = 0
...
y^4 = y^2
y^4 - y^2 = 0
y^2(y^2-1) = 0
...
1 answer