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Curve Sketching of an Exponential Function

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z overview
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}

see also:

  automatisch erstellt am 15.  6. 2016