State whether v(x) is a function or not, motivate your answer. Write down conditions that will make v a function.

A function must assign exactly one output value ($y$) for each input value ($x$).
From the graph of $v(x) = \pm\sqrt{x}$, for any $x > 0$, there are two corresponding $y$-values (one positive and one negative). For example, if $x=1$, $y=\pm 1$.
Therefore, $v(x)$ is not a function.

It fails the vertical line test because a vertical line drawn for any $x > 0$ would intersect the graph at two distinct points.
Answer: $v(x)$ is not a function because it fails the vertical line test. For a single $x$-value (where $x>0$), there are two corresponding $y$-values.

2.1.2 Write down conditions that will make $v$ a function.

To make $v(x)$ a function, we must ensure that for each $x$, there is only one $y$. This can be done by choosing either the positive or the negative branch of $\sqrt{x}$.
The conditions are:

• $v(x) = \sqrt{x}$ for $x \ge 0$ (restricting the range to $y \ge 0$).
• $v(x) = -\sqrt{x}$ for $x \ge 0$ (restricting the range to $y \le 0$). Answer: $v$ will be a function if its range is restricted to either $y \ge 0$ (i.e., $v(x) = \sqrt{x}$) or $y \le 0$ (i.e., $v(x) = -\sqrt{x}$), with the domain $x \ge 0$.

2.1.3 Determine all values of:

2.1.3.1 $y$ for which $w(x) < 0$

The condition $w(x) < 0$ means that the graph of $w(x)$ lies below the x-axis.
From the graph, $w(x) < 0$ when $0 < x < 1$.
For these $x$ values, the corresponding $y$ values (which are $w(x)$) range from $-\infty$ (as $x \to 0^+$) up to 0 (as $x \to 1^-$).
Answer: $y < \boxed{0}$ (or $y \in (-\infty, 0)$).

2.1.3.2 $x$ for which $w(x) \le -\frac{7}{10}$

We are given the point $A\left(\frac{49}{100}, -\frac{7}{10}\right)$ on the graph of $w(x)$. This means $w\left(\frac{49}{100}\right) = -\frac{7}{10}$.

Since $w(x) = \log x$ is an increasing function, if $w(x) \le -\frac{7}{10}$, then $x$ must be less than or equal to the $x$-value where $w(x) = -\frac{7}{10}$.
So, $x \le \frac{49}{100}$.
Also, the domain of $w(x) = \log x$ is $x > 0$.
Answer: $\boxed{0 < x \le \frac{49}{100}}$.

2.1.4 If a function $v(x)$ is as determined in 2.1.2 write down the equation(s) of $v^{-1}(x)$.

$v(x) = \sqrt{x}$.
Let $y = \sqrt{x}$.
To find the inverse, swap $x$ and $y$: $x = \sqrt{y}$.
Square both sides: $x^2 = y$.
The domain of $v^{-1}(x)$ is the range of $v(x)$, which is $y \ge 0$. So, $x \ge 0$.
Thus, $v^{-1}(x) = x^2$ for $x \ge 0$.

$v(x) = -\sqrt{x}$.
Let $y = -\sqrt{x}$.
To find the inverse, swap $x$ and $y$: $x = -\sqrt{y}$.
Multiply by $-1$: $-x = \sqrt{y}$.
Square both sides: $(-x)^2 = y \implies x^2 = y$.
The domain of $v^{-1}(x)$ is the range of $v(x)$, which is $y \le 0$. So, $x \le 0$.
Thus, $v^{-1}(x) = x^2$ for $x \le 0$.
Answer: The equations for $v^{-1}(x)$ are $\boxed{v^{-1}(x) = x^2 \text{ for } x \ge 0}$ and $\boxed{v^{-1}(x) = x^2 \text{ for } x \le 0}$.

2.1.5 If $h(x) = w(x) - \sqrt{x}$ where the range of $\sqrt{x}$ is $(0, \infty)$, calculate the value of $h(1)$.

Given $h(x) = w(x) - \sqrt{x}$ and $w(x) = \log x$.
So, $h(x) = \log x - \sqrt{x}$.

Substitute $x=1$ into the expression for $h(x)$:

We know that $\log(1) = 0$ and $\sqrt{1} = 1$.