Answer:
[tex]h(x')=\frac{x}{2} +\frac{2}{3}[/tex]
Step-by-step explanation:
We are given the following function:
[tex]h(x)=2x-\frac{4}{3}[/tex]
To find the inverse of this function, we will put this function equal to y (another variable) and then make x the subject of it.
[tex]y=2x-\frac{4}{3}[/tex]
[tex]y+\frac{4}{3} =2x[/tex]
[tex]\frac{y}{2} +\frac{4}{6} =x[/tex]
[tex]x=\frac{y}{2} +\frac{2}{3}[/tex]
Now changing back the variable y to x to make it the inverse of h(x) to get:
[tex]h(x')=\frac{x}{2} +\frac{2}{3}[/tex]
Therefore, the inverse of the given function h(x) is [tex]h(x')=\frac{x}{2} +\frac{2}{3}[/tex].
