The relation v(x) is defined by y2 = x. The function w(x) = rac12 an heta is defined for x>0. Consider the graph of v(x) and w(x). Analyze the given definitions and graphs to answer the following ques

2.1.1
$v(x)$ is not a function.
This is because it fails the vertical line test. For any $x > 0$, a vertical line drawn through the graph of $v(x)$ intersects it at two distinct $y$-values (one positive and one negative), which violates the definition of a function.

2.1.2
To make $v$ a function, we must restrict its range so that for each $x$-value, there is only one $y$-value.
The conditions are:

• $v(x) = \sqrt{x}$ (where $y \ge 0$)
• OR $v(x) = -\sqrt{x}$ (where $y \le 0$)

2.1.3
2.1.3.1 $y$ for which $w(x) < 0$
The function $w(x) = \log x$. From the graph, $w(x)$ is negative when $0 < x < 1$. In this interval, the $y$-values of $w(x)$ range from $-\infty$ up to (but not including) 0.
Therefore, $y < \boxed{0}$.

2.1.3.2 $x$ for which $w(x) \le -\frac{7}{10}$
The point $A\left(\frac{49}{100}; -\frac{7}{10}\right)$ is on the graph of $w(x)$. Since $w(x) = \log x$ is an increasing function, for $w(x)$ to be less than or equal to $-\frac{7}{10}$, the corresponding $x$-values must be less than or equal to $\frac{49}{100}$. Also, the domain of $w(x) = \log x$ is $x > 0$.
Therefore, $\boxed{0 < x \le \frac{49}{100}}$.

2.1.4
Assuming "5.1.2" is a typo and refers to "2.1.2", we consider the two conditions for $v(x)$ to be a function:

• Case 1: If $v(x) = \sqrt{x}$ (with domain $x \ge 0$ and range $y \ge 0$)