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Find any stationary points on the curve 3x^2 +x^2y +y= 2?

+3 votes

show working please

asked Feb 12, 2013 in PRECALCULUS by angel12 Scholar

1 Answer

+4 votes

3x2 + x2y + y = 2.------------------>(1).

Apply derivative each side.

(d/dx)(3x2 + x2y + y) = (d/dx)2.

(d/dx)(3x2) + (d/dx)(x2y) + (d/dx)(y) = (d/dx)2.

Power Rule of Derivative: (d/dx)(xn)=nxn-1 and Product Rule: (UV)'=U'V+UV'

3(2x) + 2xy + x2y' + y' = 0

6x + 2xy + y'(x2 +1) = 0

The stationary point on the curve is y' = 0 but y not equal to 0.

So, 6x + 2xy = 0

Take out common term x.

x(6 + 2y) = 0

Here x = 0

Substitute x = 0 in the equation (1).

3(0)2 + (0)2y + y = 2

So, y = 2

Therefore the stationary point is (x, y)=(0, 2).

answered Feb 12, 2013 by richardson Scholar

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