We are given four problems: a) i. Use the vertical line test to show that $f(x) = \sqrt{x}$ is a function. ii. Determine the range that makes $f(x) = \sqrt{x}$ a function. b) Describe the effect of adding a constant to the function $f(x) = 3x^2$. c) Given the linear function $f(x) = ax + b$ with $f(1) = 8$ and $f(3) = 14$, find the values of $a$ and $b$. d) For the function $g(x) = \frac{x+1}{x-1}$, find the inverse function $g^{-1}(x)$ and determine the value of $x$ for which $g^{-1}(x)$ is not defined.

1. Problem Description

We are given four problems:

a) i. Use the vertical line test to show that $f(x) = \sqrt{x}$ is a function.

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

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

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

d) For the function $g(x) = \frac{x+1}{x-1}$, find the inverse function $g^{-1}(x)$ and determine the value of $x$ for which $g^{-1}(x)$ is not defined.

2. Solution Steps

a) i. Vertical Line Test: If any vertical line intersects the graph of a relation more than once, then the relation is not a function. The graph of $f(x) = \sqrt{x}$ is the top half of a parabola opening to the right. Any vertical line $x = c$ will intersect the graph at most once for $c \ge 0$. Thus, $f(x) = \sqrt{x}$ is a function.

ii. The range of $f(x) = \sqrt{x}$ is $[0, \infty)$.

b) Adding a constant $c$ to the function $f(x) = 3x^2$ gives $f(x) = 3x^2 + c$. 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 $f(3) = 14$.

Substituting $x=1$ into $f(x)$, we get $f(1) = a(1) + b = a + b = 8$.

Substituting $x=3$ into $f(x)$, we get $f(3) = a(3) + b = 3a + b = 14$.

We have a system of two linear equations with two variables:

$a + b = 8$

$3a + b = 14$

Subtracting the first equation from the second equation, we get:

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

$2a = 6$

$a = 3$

Substituting $a=3$ into $a + b = 8$, we get $3 + b = 8$, so $b = 5$.

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

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

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

$xy - x = y + 1$

$xy - y = x + 1$

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

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

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

The inverse function $g^{-1}(x)$ is not defined when the denominator is zero, i.e., $x-1=0$. Thus $x=1$.

3. Final Answer

a) i. The vertical line test shows that $f(x) = \sqrt{x}$ is a function.

ii. The range is $[0, \infty)$.

b) Adding a constant shifts the graph vertically.

c) $a = 3$ and $b = 5$.

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