I've tried differentiating this over again, and took me a while to get it. but Im not sure, my answer were (1/3, 4/3)
heres the question
At what point on the curve xy=(1-x-y)^2 is the tangent line parallel to the x-axis.
Thank you very much
I've tried differentiating this over again, and took me a while to get it. but Im not sure, my answer were (1/3, 4/3)
heres the question
At what point on the curve xy=(1-x-y)^2 is the tangent line parallel to the x-axis.
Thank you very much
wow. okay thanks..but okay to tell you the truth, i didn't actually solve the problem i just saw the answer at the back of the book, I know the process differentiate it then set the dy/dx zero since parallel means same slope and x-axis is is horizontal meaning slope is zero.. but what i encountered is I don't really know how to get the rest of the point...i tried substituting the y to x in the dy/dx formula but i can never separate y from x... so there's 2 integers on dy/dx=-(y-2-2x)/(2+2x+4y)
Implicitly differentiating, you should find:
$\displaystyle \frac{dy}{dx}=\frac{2x+y-2}{2-x-2y}$
Equating this to zero implies:
$\displaystyle y=2(1-x)$
Substituting this into the original equation gives:
$\displaystyle -2x(x-1)=(x-1)^2$
$\displaystyle (x-1)(3x-1)=0$
$\displaystyle x=\frac{1}{3},1$
So the two points are: $\displaystyle \left(\frac{1}{3},\frac{4}{3} \right),(1,0)$