1. ## Inverse functions help?

the questino is
A)find domain and range
b)find domain and range of inverse function
c)Find an explicit formula for the inverse function as a function of x.
$\displaystyle f(x) (x-2)^2-9 x is bigger than or equal to 4$

a)domain of original function is [4, infinity)
How do I find the range?
b)domain of inverse function is range of original function, range of inverse function is domain of original function so [4, infinity)
c) the formula for the inverse function is
$\displaystyle f(x)=(x-2)^2-9$
$\displaystyle x=(f(x)-2)^2-9$
$\displaystyle x+9 = (f(x)-2)^2$
$\displaystyle \sqrt{x+9} = f(x)-2$
$\displaystyle 2+\sqrt{x+9} = f(x)$
i would think that the range would be [9, infinity), since 9 would be the lowest number than can still make the square root non negative, but the book's answer says the range of the original function/domain of inverse is [-5, infinity) does this have something to do with the restricted domain? if so, what do I do to get this range?

2. I like the work you have done here. It might be useful to know that

the domain of $\displaystyle f(x)$ is the range of $\displaystyle f'(x)$ and

the range of $\displaystyle f(x)$ is the domain of $\displaystyle f'(x)$

To make your square roots look better use \sqrt{} instead of \sqrt()

This is the difference between $\displaystyle \sqrt{x+9}$ and $\displaystyle \sqrt(x+9)$

3. Thanks for the tip about the square roots, i couldn't figure out how to make it look right.

the problem is I don't know how to find the range of the original f(x), so I can't find the domain of f'(x)

4. Graph the function to get a rough idea of the shape and turning point.

Here's a hint, the range for your function is $\displaystyle [f(4),\infty)$

Now find $\displaystyle f(4)$

5. f(4) = -5
why do I plug in the 4 to get the range?

6. [HTML][/HTML]
Originally Posted by dorkymichelle
f(4) = -5
why do I plug in the 4 to get the range?
The point of using "4" is that your domain is $\displaystyle [4, \infty)$.

Another important point is that $\displaystyle f(x)= (x-2)^2- 9$ is equal to -9 when x= 2 and is larger than -9 for all other x, because $\displaystyle (x- 2)^2$ is positive for all x except 2. That is, the graph is a parabola with vertex at (2, 9). But your domain is $\displaystyle x\ge 4$- what does the parabola look like past x= 4? What does that tell you about the domain?

Now, as for the inverse, a standard way of finding inverse functions is to "swap" x and y: if y= f(x) then $\displaystyle x= f^{-1}(y)$. From $\displaystyle y= (x-2)^2- 9$ you get $\displaystyle x= (y- 2)^2- 9$. Solve for y. When taking the square root, to 'invert' the square, remember that y must be larger than or equal to 4.