Answer:
[tex]x=\frac{(y-4)^2}{6}+\frac{17}{9}[/tex]
is the inverse of the function.
Step-by-step explanation:
We have been given the function:
[tex](x-4)^2-\frac{2}{3}=6y-12[/tex]
For inverse function we will swap x and y
And firstly, we will find the value of y in terms of x
and then replace x and y
[tex]6y=(x-4)^2-\frac{2}{3}+12[/tex]
[tex]6y=\frac{3(x-4)^2-1+36}{3}[/tex]
[tex]y=\frac{(x-4)^2}{6}+\frac{17}{9}[/tex]
Now, we will replace x and y
[tex]x=\frac{(y-4)^2}{6}+\frac{17}{9}[/tex]
Answer:
[tex]y=\sqrt{6y-\frac{34}{3}}+4[/tex]
Step-by-step explanation:
[tex](x-4)^2-\frac{2}{3}=6y-12[/tex]
In order to find the inverse of above function, we solve it for x in term of y and make the replace x with y in the final answer.
Adding \frac{2}{3} on both hand sides we ger
[tex] (x-4)^2=6y-12+\frac{2}{3}\\(x-4)^2=6y+\frac{-36+2}{3}\\(x-4)^2=6y+\frac{-34}{3}[/tex]
[tex](x-4)^2=6y-\frac{34}{3}[/tex]
Taking square root on both hand sides we get
[tex] (x-4)=\sqrt{6y-\frac{34}{3}}[/tex]
Adding 4 on both sides we get
[tex]x=\sqrt{6y-\frac{34}{3}}+4[/tex]
Replacing x with y and y with x we get
[tex]y=\sqrt{6y-\frac{34}{3}}+4[/tex]
