SOLUTION: Find the domain and range of the inverse of the function. f(x) = 6-x^2; x is greater than or equal to 0.

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Question 169749: Find the domain and range of the inverse of the function.
f(x) = 6-x^2; x is greater than or equal to 0.

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Find the domain and range of the inverse of the function.

matrix%281%2C3%2C+++f%28x%29+=+6-x%5E2%2C%22%3B%22%2C+x+%3E=+0%29  

Let's do it graphically first, and then we'll do it
algebraically.

First we'll draw the graph of y=6-x%5E2



That is a parabola with vertex (0,6).
However because of the x%3E=0
restriction, we will have to eliminate
the left half of the graph to have the
graph of f(x), which is this:



The domain of f%28x%29 is matrix%281%2C5%2C+%22%5B%22%2C+0%2C+%22%2C%22%2C+infinity%2C+%22%29%22%29   
The range of f%28x%29 is matrix%281%2C5%2C+%22%28%22%2C+-infinity%2C+%22%2C%22%2C+6%2C+%22%5D%22%29

Now let's draw what is called "the identity line",
which is the 45°-line through the origin whose 
equation is y=x.  I'll draw the identity line
in green dotted:



Now let's draw the graph of matrix%281%2C2%2Cf%5E%28-1%29%2C%28x%29%29,
the inverse of f%28x%29 by reflecting it across the 
green dotted identity line like this:



The domain of f%5E%28-1%29%28x%29 is matrix%281%2C5%2C+%22%28%22%2C+-infinity%2C+%22%2C%22%2C+6%2C+%22%5D%22%29
The range of f%5E%28-1%29%28x%29 is matrix%281%2C5%2C+%22%5B%22%2C+0%2C+%22%2C%22%2C+infinity%2C+%22%29%22%29


Notice that the range of f%28x%29 is the domain of f%5E%28-1%29%28x%29,
and the domain of f%28x%29 is the range of f%5E%28-1%29%28x%29.
That's doing it graphically. Now let's do it
algebraically:

matrix%281%2C3%2C+++f%28x%29+=+6-x%5E2%2C%22%3B%22%2C+x+%3E=+0%29

Replace f%28x%29 by y

matrix%281%2C3%2C+++y+=+6-x%5E2%2C%22%3B%22%2C+x+%3E=+0%29

Interchange x and y:

matrix%281%2C3%2C+++x+=+6-y%5E2%2C%22%3B%22%2C+y+%3E=+0%29

Solve for y:

matrix%281%2C3%2C+++y%5E2+=+6-x%2C%22%3B%22%2C+y+%3E=+0%29

We take the square roots of both sides, and 
since y%3E=0, we do not need "±".

matrix%281%2C3%2C+++y+=+sqrt%286-x%29%2C%22%3B%22%2C+y+%3E=+0%29

Now we replace y by f%5E%28-1%29%28x%29



What is under the square root radical must not be negative,
so 6-x%3E=0 or -x%3E=-6 or x%3C=6

So, because of x%3C=6,

The domain of f%5E%28-1%29%28x%29 is matrix%281%2C5%2C+%22%28%22%2C+-infinity%2C+%22%2C%22%2C+6%2C+%22%5D%22%29

and because of the y%3E=0,

The range of f%5E%28-1%29%28x%29 is matrix%281%2C5%2C+%22%5B%22%2C+0%2C+%22%2C%22%2C+infinity%2C+%22%29%22%29

Edwin