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FM Nov 2023 p12 q07
4191
The curve C has equation \(y = f(x)\), where \(f(x) = \frac{x^2}{x+1}\).
Find the equations of the asymptotes of C.
Find the coordinates of any stationary points on C.
Sketch C.
Find the coordinates of any stationary points on the curve with equation \(y = \frac{1}{f(x)}\).
Sketch the curve with equation \(y = \frac{1}{f(x)}\) and find, in exact form, the set of values for which \(\frac{1}{f(x)} > f(x)\).
Solution
(a) The vertical asymptote occurs where the denominator is zero, so \(x = -1\). For the oblique asymptote, perform polynomial long division on \(\frac{x^2}{x+1}\) to get \(y = x - 1\).
(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate \(y = \frac{x^2}{x+1}\) to get \(\frac{dy}{dx} = \frac{x^2 + 2x}{(x+1)^2}\). Solve \(x^2 + 2x = 0\) to find \(x = 0\) and \(x = -2\). Substitute back to find \(y\) values: \((0, 0)\) and \((-2, -4)\).
(c) Sketch the curve using the asymptotes and stationary points. The curve approaches the asymptotes and passes through the stationary points.
(d) For \(y = \frac{1}{f(x)}\), differentiate and set \(\frac{dy}{dx} = 0\) to find stationary points. The stationary point is \((-2, -\frac{1}{4})\).
(e) Sketch \(y = \frac{1}{f(x)}\) using the asymptotes and stationary points. Solve \(\frac{1}{f(x)} > f(x)\) by setting \(\frac{x^2}{x+1} = 1\) and \(\frac{x^2}{x+1} = -1\). Solve \(x^2 - x - 1 = 0\) to find critical points \(x = \frac{1}{2} \pm \frac{1}{2} \sqrt{5}\). The solution set is \(x < -1, -\frac{1}{2} \sqrt{5} < x < \frac{1}{2} \sqrt{5}, x \neq 0\).
