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Mathematics-Online course: Preparatory Course Mathematics - Analysis - Extrema and Curve Sketching

Curve Sketching

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In order to describe the qualitative behavior of a function, the following properties can be considered:

\includegraphics[width=14.5cm]{Kurvendiskussion_en.eps}
We analyze the function $ f(x)=\sin(x) + \frac{1}{3}\,\sin(3x)$.

Symmetry: Since $ \sin(x)=-\sin(-x)$, the function is odd.

Periodicity: As the sine function, the function is $ 2\pi$-periodic. Hence, it is sufficient to consider the interval $ [-\pi,\pi]$.

Points of discontinuity and Poles: The function is a composition of continuous functions, and, therefore, has no points of discontinuity or poles.

Zeros: By the addition theorem, $ \sin(3x) = 3\sin(x)-4\sin^3(x)$, and it follows that

$\displaystyle f(x) = 2 \sin(x)-\frac{4}{3}\,\sin^3(x) = \sin(x)\underbrace{\left(2-\frac{4}{3}\sin^2(x)\right)}_{\neq 0} \,. $

Hence, $ f$ vanishes at 0 and $ \pm \pi$.

Extrema: The derivative

$\displaystyle f^\prime(x) = 2 \cos(x)-4\sin^2(x)\cos(x) = -2\cos(x)+4\cos^3(x) $

vanishes if $ \cos(x) = 0$ or $ \cos(x)=\pm 1/\sqrt{2}$. Consequently, the critical points of $ f$ are

$\displaystyle x=\pm \pi/2 \,,\quad x=\pm \pi/4 \,,\quad x= \pm 3\pi/4 \,. $

Since the function is periodic and defined on $ \mathbb{R}$, we do not have to consider boundary values. Thus, the type of the critical points can be determined from the sign of the second derivative

$\displaystyle f^{\prime\prime}(x) = 2 \sin x -12 \cos^2 x \, \sin x = 2 \sin x(1-6\cos^2 x) $

and by comparing function values. This yields

\begin{displaymath} \begin{array}{c\vert c\vert c\vert l} x & f(x) & f^{\prime\... ...& \sqrt{8}/3 & -2\sqrt{2}<0 & \mbox{global maximum} \end{array}\end{displaymath}

Inflection Points: The second derivative

$\displaystyle f^{\prime\prime}(x) = 2\sin x -12\cos^2 x \,\sin x=\sin x\,(6\sin^2x-5) $

vanishes if $ \sin x =0$ or $ \sin x =\pm \sqrt{5/6}$, i.e.

$\displaystyle x=0 \quad\lor\quad x=\pm \pi \quad\lor\quad x=\pm a \quad\lor\quad x = \pm(\pi-a) $

with $ a = \operatorname{arcsin}\sqrt{5/6}\approx {\tt 1.1503}$. Since the third derivative is nonzero at these points, $ f$ has inflection points at $ (0,0)$, $ (\pm \pi,0)$, $ (\pm a,\pm b)$, and $ (\pm(\pi-a),\pm b)$ with

$\displaystyle b = f(a) = \sin a\,(2-(4/3)\sin^2 a) = \sqrt{5/6}\,(2-(4/3)(5/6)) \approx {\tt0.8114} \,, $

noting that $ \sin a = \sin(\pi-a)$ and $ f(-a)=-f(a)$.

Asymptots: Since $ f$ is periodic and not constant, it has no asymptotes.

\includegraphics[width=10.4cm]{Kurvendiskussion_1_en}

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}

We analyze the function

$\displaystyle f(x) = \vert x^2-1\vert$e$\displaystyle ^{-4x/3} \,. $

(i) Qualitative behavior: As the exponential function, $ f$ does not possess any symmetries and is not periodic.

The derivative is discontinuous at $ x=\pm 1$ in view of the discontinuity of the absolute value at the argument 0. Since $ \lim_{x\to\infty}x^r\exp(-sx) = 0$ for all $ r,s>0$, $ p(x) = 0$ is the asymptote to $ f$ for $ x\to\infty$. For $ x\to-\infty$ an asymptote does not exist since $ \lim_{x\to-\infty}\vert f(x)/x\vert=\infty$.

(ii) Zeros: In view of the positivity of the exponential function, the zeros of $ f$ are determined by the first factor und equal $ x_{1,2}=\pm 1$. Since $ f\ge0$, the zeros are also global minima. A global maximum does not exist since $ \lim_{x\to-\infty}f(x) = \infty$.

(iii) Extrema: Since $ f(-1) = f(1) = f(\infty) = 0$, the intervals $ (-1,1)$ and $ (1,\infty)$ each contain at least one local maximum. Differentiating

$\displaystyle f(x) = \sigma (x^2-1)$e$\displaystyle ^{-4x/3},\quad x\ne\pm1\,, $

with $ \sigma = -1$ for $ x\in(-1,1)$ and $ \sigma = 1$ for $ x\in(-\infty,-1)\cup(1,\infty)$, yields

$\displaystyle f^\prime(x) = \sigma \left[-4x^2/3+4/3+2x\right]$e$\displaystyle ^{-4x/3} \,. $

Setting the quadratic polynomial in brackets to zero, we obtain the critical points $ x_3=-1/2$ und $ x_4=2$. At each point $ f$ has a local maximum in view of the existence of at least two such extrema. The corresponding function values are

$\displaystyle y_3 = \frac{3}{4}$e$\displaystyle ^{-2/3} \approx {\tt 1.4608}, \quad y_4 = 3$e$\displaystyle ^{-8/3} \approx {\tt0.2085} \,. $

(iv) Inflection points: The zeros of

$\displaystyle f^{\prime\prime}(x) = \sigma\left[16x^2/9-16x/3+2/9\right]$e$\displaystyle ^{-4x/3} \,, $

i.e., of $ [\ldots]$, are

$\displaystyle x_5 = 3/2-\sqrt{34}/4 \approx {\tt0.0423},\quad x_6 = 3/2+\sqrt{34}/4 \approx {\tt 2.9577} \,. $

Both are inflection points since $ f^{\prime\prime}$ changes sign. The function values are

$\displaystyle y_5 \approx {\tt0.9435},\quad y_6 \approx {\tt0.1501} \,. $

\includegraphics[width=10.4cm]{Kurvendiskussion_3_en}
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