1. Oct 27, 2007

### STAR3URY

I don't understand this at all...

Find the Range and Inverse of f(x) = square root of [(8x^2 +1)/ (9-5x^2)]

2. Oct 27, 2007

### Gib Z

The range of a function is all the values f(x) can take. The inverse of a function is basically the function that cancels it out. The inverse of division by 3 is multiplication by 3. The inverse of square rooting a positive number is squaring that number. To find the inverse function when you have a general $$y=f(x)$$ is to swap the y's and x's, ie $$x=f(y)$$, and it is not necessarily required to solve again for y, and in general it is not even possible in terms of elementary functions, but sometimes you lose marks when you can but you don't.

3. Oct 29, 2007

### andytoh

The range is easily solvable because only x^2 terms exist. Solving for x in terms of y can thus be done by a simple square root. Just make sure the radicand is not negative and that will give you the restrictions for the range y.