Consider the curve given by the equation x^3+xy=-2 It can be shown that (dy)/(dx)=(-3x^2-y)/(x) Find the point on the curve where the line tangent to the curve is horizontal. d square

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Consider the curve given by the equation x^3+xy=-2 It can be shown that (dy)/(dx)=(-3x^2-y)/(x) Find the point on the curve where the line tangent to the curve is horizontal. d square

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Answer 1

To find the point on the curve where the line tangent to the curve is horizontal, we need to determine where the derivative \(\frac{dy}{dx}\) is equal to zero. The given curve is described by the equation:\[x^{3} + xy = -2\]And the derivative is given by:\[\frac{dy}{dx} = \frac{-3x^{2} - y}{x}\]A horizontal tangent line occurs where the slope of the tangent line is zero. This happens when the numerator of the derivative is zero, since a fraction is zero when its numerator is zero (and its denominator is not zero).So, we set the numerator of the derivative equal to zero and solve for \(x\) and \(y\):\[-3x^{2} - y = 0\]Solving for \(y\) gives us:\[y = -3x^{2}\]Now, we need to find(s) of \(x\) for which this solution is valid. We substitute \(y = -3x^{2}\) back into the original equation to find the corresponding \(x\) values:\[x^{3} + x(-3x^{2}) = -2\]\[x^{3} - 3x^{3} = -2\]\[-2x^{3} = -2\]\[x^{3} = 1\]\[x = 1\]Substituting \(x = 1\) back into the equation \(y = -3x^{2}\) gives us:\[y = -3(1)^{2}\]\[y = -3\]Therefore, the point on the curve where the line tangent to the curve is horizontal is \((1, -3)\).

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Consider the curve given by the equation x^3+xy=-2 It can be shown that (dy)/(dx)=(-3x^2-y)/(x) Find the point on the curve where the line tangent to the curve is horizontal. d square

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