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Curve Sketching of a Rational Function

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We analyze the function

$\displaystyle f(x)=\frac{5x^3+4x}{x^2-1} \,. $

Symmetry: The numerator is even and the denominator is odd. Thus the function is odd, i.e., symmetric with respect to the origin.

Periodicity: The function is not periodic.

Points of Discontinuity and Poles: The denominator of $ f$ has simple zeros at $ x=\pm 1$. Since the numerator is nonzero at these points, the singularities are not removable and hence $ -1$ and $ 1$ are simple poles.

Zeros: The nominator vanishes at $ x=0$.

Extrema: The derivative

$\displaystyle f^\prime(x) = \frac{5x^4-19x^2-4}{(x^2-1)^2} = \frac{(x^2-4)(5x^2+1)}{(x^2-1)^2} $

vanishes at $ x = \pm 2$. The type of these critical points can be determined from the qualitative behavior of $ f$. First we observe that $ f$ does not possess global extrema in view of the simple poles where the sign changes. Moreover, since $ f(x)\to-\infty$ for $ x\to -\infty$ and $ x\to -1$, the interval $ \left(-\infty,-1 \right)$ must contain a local maximum. Similarly, $ \left( 1, \infty \right)$ contains at least one local minimum. Hence, the zeros of $ f^\prime$ both correspond to local extrema, a local maximum at $ (-2,-4/5)$ and a local minimum at $ (2,4/5)$.

Inflection Points: The second derivative

$\displaystyle f^{\prime\prime}(x) = \frac{18x(x^2+3)}{(x^2-1)^3} $

has a unique zero with sign change at $ x=0$. Hence, $ (0,0)$ is an inflection point of $ f$.

Asymptotes: Polynomial division yields

$\displaystyle f(x) = 5x + 0 + \frac{9x}{x^2-1} $

and the asymptote $ p(x)=5x$.

\includegraphics[width=10.4cm]{Kurvendiskussion_2_en}

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