pathterminuspages/math/aboutcontactabout me

3.The Square Root



1. Logarithmic
2. Absoulte Value
@3. The Square Root

The square root of $x$ is written as a function as $$ f(x) = \sqrt{x} $$ It is the inverse of $g(x) = x^2$, that is $g \circ f = f \circ g = id$. Or we can state this as $$ \sqrt{x^2} = \sqrt{x}^2 = x $$ The square root is defined in the range $[0,\infty]$, thus not for $x \lt 0$. We have for $x \lt 0$, that $x^2 \gt 0$. So though $f \circ g$ is defined on $[-\infty,\infty]$, we have that this compound function is not injective. It can be surjective, if defined properly: $$ f \circ g : \mathbb{R} \rightarrow \mathbb{R}^{+} $$

We can plot the function with the following Python3 code:

import numpy as np import matplotlib.pyplot as plt def f(x): return np.sqrt(x) x = [float(x) / 10.0 for x in range(0,10 ** 4)] y = [f(x) for x in x] plt.plot(x,y)

For which we get the plot as seen in Figure 1.

Figure 1: plot of the absolute value $|x|$.

The derivative of the absolute value

The derivative is given as $$ f(x)' = \frac{1}{2 \sqrt{x}} $$ We can show this as follows. Rewrite $$ \sqrt{x} = x^{1/2} $$ Use the power rule to obtain $$ f(x)' = \frac{1}{2} x^{-1/2} $$ We have that $$ x^{-1/2} = \frac{1}{x^{1/2}} = \frac{1}{\sqrt{x}} $$ So in total we have that $$ f(x)' = \frac{1}{2} \cdot \frac{1}{\sqrt{x}} = \frac{1}{2 \sqrt{x}} $$

CommentsGuest Name:Comment: