Question

a) determine the inverse of the function by interchanging the variables and solving for y in terms of x.
$y = \frac{1}{2}x - \frac{3}{2}$
options:
$y = 2x + 3$
$y = 3x - 2$
$y = -3x - 2$
$y = 2x - \frac{2}{3}$

b) determine the inverse of the function by interchanging the variables and solving for y in terms of x.
$y = 4x^2$
options:
$y = \pm\sqrt{x}$
$y = \pm 2x$
$y = \pm \frac{\sqrt{x}}{2}$
$y = \pm 4\sqrt{x}$

c) determine the inverse of the function by interchanging the variables and solving for y in terms of x.
$y = x^2 - 1$
options:
$y = 1 \pm \sqrt{x}$
$y = -1 \pm \sqrt{x}$
$y = \pm \sqrt{x + 1}$
$y = -x^2 + 1$

Explanation:

Step1: Invert linear function

Start with $y=\frac{1}{2}x-\frac{3}{2}$. Swap $x$ and $y$:
$x=\frac{1}{2}y-\frac{3}{2}$
Solve for $y$:
$x+\frac{3}{2}=\frac{1}{2}y$
$y=2x+3$

Step2: Invert quadratic function 1

Start with $y=4x^2$. Swap $x$ and $y$:
$x=4y^2$
Solve for $y$:
$y^2=\frac{x}{4}$
$y=\pm\frac{\sqrt{x}}{2}$

Step3: Invert quadratic function 2

Start with $y=x^2-1$. Swap $x$ and $y$:
$x=y^2-1$
Solve for $y$:
$y^2=x+1$
$y=1\pm\sqrt{x}$

Answer:

1. $y = 2x + 3$
2. $y = \pm\frac{\sqrt{x}}{2}$
3. $y = 1\pm\sqrt{x}$