We are given four problems: a) i. Show that $f(x) = \sqrt{x}$ is not a function using the vertical line test, which is incorrect. ii. Determine the range that makes $f(x) = \sqrt{x}$ a function. b) Determine the effect of adding a constant to the function $f(x) = 3x^2$. c) Given the linear function $f(x) = ax + b$ such that $f(1) = 8$ and $f(3) = 14$, find the values of $a$ and $b$. d) The function $g(x) = \frac{x+1}{x-1}$ is defined for the domain $\{x \in \mathbb{R}: x \neq 1\}$. Find its inverse function, $g^{-1}(x)$, and state the value for $x$ for which $g^{-1}(x)$ is not defined.

1. Problem Description

We are given four problems:

a) i. Show that $f(x) = \sqrt{x}$ is not a function using the vertical line test, which is incorrect.

ii. Determine the range that makes $f(x) = \sqrt{x}$ a function.

b) Determine the effect of adding a constant to the function $f(x) = 3x^2$.

c) Given the linear function $f(x) = ax + b$ such that $f(1) = 8$ and $f(3) = 14$, find the values of $a$ and $b$.

d) The function $g(x) = \frac{x+1}{x-1}$ is defined for the domain $\{x \in \mathbb{R}: x \neq 1\}$. Find its inverse function, $g^{-1}(x)$, and state the value for $x$ for which $g^{-1}(x)$ is not defined.

2. Solution Steps

a) i. The function $f(x) = \sqrt{x}$ *is* a function. The vertical line test states that a graph represents a function if any vertical line drawn through the graph intersects it at most once. The graph of $f(x) = \sqrt{x}$ passes the vertical line test since for any $x \geq 0$, there is only one value of $f(x)$. The problem is incorrect in saying it is not a function.

ii. The range of $f(x) = \sqrt{x}$ is all non-negative real numbers, i.e., $f(x) \geq 0$. So, the range is $[0, \infty)$.

b) Adding a constant $c$ to the function $f(x) = 3x^2$ results in the new function $f(x) = 3x^2 + c$. This shifts the graph of $f(x) = 3x^2$ vertically by $c$ units. If $c > 0$, the graph shifts upward; if $c < 0$, the graph shifts downward.

c) We are given $f(x) = ax + b$, $f(1) = 8$, and $f(3) = 14$. We can write two equations:

$f(1) = a(1) + b = a + b = 8$

$f(3) = a(3) + b = 3a + b = 14$

Subtracting the first equation from the second, we get:

$(3a + b) - (a + b) = 14 - 8$

$2a = 6$

$a = 3$

Substituting $a = 3$ into the first equation, we get:

$3 + b = 8$

$b = 5$

Thus, $a = 3$ and $b = 5$.

d) Let $y = g(x) = \frac{x+1}{x-1}$. To find the inverse function, we swap $x$ and $y$ and solve for $y$:

$x = \frac{y+1}{y-1}$

$x(y-1) = y+1$

$xy - x = y + 1$

$xy - y = x + 1$

$y(x-1) = x+1$

$y = \frac{x+1}{x-1}$

So, $g^{-1}(x) = \frac{x+1}{x-1}$.

The inverse function $g^{-1}(x)$ is undefined when the denominator is zero, i.e., when $x-1 = 0$, which means $x = 1$.

3. Final Answer

a) i. $f(x) = \sqrt{x}$ is a function and passes the vertical line test.

ii. Range: $[0, \infty)$

b) Adding a constant $c$ to $f(x) = 3x^2$ shifts the graph vertically by $c$ units.

c) $a = 3$, $b = 5$

d) $g^{-1}(x) = \frac{x+1}{x-1}$. $g^{-1}(x)$ is not defined for $x = 1$.