Recent content by ClaudeFrollo

I thought I answered this when I said ((-x2(4-x2))/(\sqrt{4-x^{2}}) ...wouldn't the numerator just be ((-x2(4-x2))?

when x= -2, 2?

I meant -x^2, not 1-x D:

Ohhhhh yeah, so I guess ((1-x)+(4-x^{2}))/(\sqrt{4-x^{2}}) so f'(0)= 5/4?

Whoops. I think I meant: x(\sqrt{4-x^{2}})/(\sqrt{4-x^{2}}) so... y'=-x? So x=0 when y'=0?

(1-x)/\sqrt{4-x^{2}}=y' y'= 1/2 Hmm... so does that mean that that's the slope of the horizontal tangent line?

sorry, i hit enter too soon Homework Statement Y=x(sqrt(4-x^2)) how do I find all pts on the graph where the tan line is vert and hori? Homework Equations Y=x(sqrt(4-x^2)) The Attempt at a Solution y'=(x)(.5(4-x^2)^(-1/2)(-2x))+(sqrt(4-x^2))(1)