Question

find the inverse of $f(x)=\frac{2}{3}x - \frac{1}{3}$. the inverse is $g(x)=square$. graph the function and its inverse.

Explanation:

Step1: Replace $f(x)$ with $y$

$y = \frac{2}{3}x - \frac{1}{3}$

Step2: Swap $x$ and $y$

$x = \frac{2}{3}y - \frac{1}{3}$

Step3: Isolate the term with $y$

Add $\frac{1}{3}$ to both sides:
$x + \frac{1}{3} = \frac{2}{3}y$

Step4: Solve for $y$

Multiply both sides by $\frac{3}{2}$:
$y = \frac{3}{2}x + \frac{3}{2} \times \frac{1}{3}$
Simplify the constant term:
$y = \frac{3}{2}x + \frac{1}{2}$

Step5: Rename $y$ as $g(x)$

$g(x) = \frac{3}{2}x + \frac{1}{2}$

Graphing Notes:

• For $f(x)=\frac{2}{3}x - \frac{1}{3}$: y-intercept at $(0, -\frac{1}{3})$, x-intercept at $(\frac{1}{2}, 0)$, slope $\frac{2}{3}$.
• For $g(x)=\frac{3}{2}x + \frac{1}{2}$: y-intercept at $(0, \frac{1}{2})$, x-intercept at $(-\frac{1}{3}, 0)$, slope $\frac{3}{2}$.
• The two lines are symmetric across $y=x$.

Answer:

$g(x) = \frac{3}{2}x + \frac{1}{2}$