hello
am stopped at this problem -_-
let where x>2
Find a formula for the inverse function of it, if exists.
My solution:
f`(x)=4x-..... so f is increas..so its 1-1funct..so inverse is exist.
its easy to prove that the inverse function is exist..=)
But the problem is finding the inverse function
ok let ..
how to find the x in terms of y !!
i tried to get the natural logarthmic of both sides but it failed :S
any help?!
IS one to one, but only because the domain has been restricted to all x-values greater than the turning point, not for the reason you stated.
The domain of f is , so the range is , since the function is increasing when ...
Note that for an inverse function, the domain and range swap, which implies that the x and y values swap.
So
Can you solve for y now? Remember that the domain of is since .
Use the standard process:
Rename "f" as "y": y = 2x^2 - 8x
Solve this for "x=". In this case, use the Quadratic Formula:
. . . . .
. . . . .
...and so forth. In this case, you were given the restriction of x > 2, so you know you need only the positive square root:
. . . . .
Then continue with the usual steps: Swap "x" and "y", and rename the new "y" as f-inverse.