Graphing error found

Question

$f(x)=e^x|\ln(x)|$ is defined over $(0,+\infty)$.

How is there a negative part of the function shown by graphing it?!

Answer 1

You're right that $$f(x) = e^x|\log(x)|$$ is only defined on $(0,+\infty$), so the "problem" lies with the graphing software.

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Below is a plausible explanation that seems to match the graph drawn. Many calculators and software follow the steps listed elow, e.g. my TI-83, WolframAlpha and GeoGebra.

Step 1: interpreting the logarithm of a negative real number.

The logarithm on the reals is not defined when $x \leq 0$, but one does have the principal value of the complex logarithm: $$ \operatorname{Log}: \mathbb{C}\setminus \{0\} \to \mathbb{C}: z = re^{i\theta} \mapsto \log(r) + i\theta,$$ where $r > 0$ and $\theta \in (-\pi,\pi]$.

In particular, for a real $x < 0$ you have $$ \operatorname{Log}(x) = \operatorname{Log}(|x|e^{\pi i}) = \log(|x|) + i\pi.$$ Calculators and software that can calculate this complex logarithm, will often default to using this when asked to calculate the logarithm of a negative number.

Step 2: interpreting the absolute value.

While the absolute value $|\;.\,|$ is defined on the real numbers, there is natural extension to the complex numbers: the modulus function $$|\;.\,|: \mathbb{C} \to \mathbb{R}^+: z = a+bi \mapsto \sqrt{a^2+b^2}.$$ Again, many calculators and software will default to this function when asked to calculate the absolute value of a nonreal number.

In the case of $\operatorname{Log}(x)$ with $x < 0$ a real number, we have: $$|\operatorname{Log}(x)| = |\log(|x|) + i\pi| = \sqrt{\log(|x|)^2 + \pi^2}.$$

Result:

The function $$f: (0,+\infty) \to \mathbb{R}: x \mapsto e^x |\log (x)|$$ gets extended to the function $$f: \mathbb{R}\setminus \{0\} \to \mathbb{R}: x \mapsto \left\{\begin{array} {ll} e^x |\log (x)| &\text{ if } x > 0\\ e^x \sqrt{\log(|x|)^2 + \pi^2} &\text{ if } x < 0 \end{array}\right.,$$ which is what you get on your graphing app.