already figured out the domain and range, I know that I have to choose the positive
switch the x's and y's;
the new "y
=" is the inverse:
Here's the graph:
Then the inverse is y
= sqrt(x – 1), x>
and the inverse is also a function.
If you've studied function notation,
you may be starting with "f(x)"
instead of "y". In that case, start the inversion process by renaming
"y"; find the inverse, and rename the resulting "y" as "f–1(x)".
It's usually easier to work with "y". Warning: This notation is misleading; the "minus
one" power in the function notation means "the inverse function", not "the
reciprocal of". Don't confuse the two.
Find the inverse of y = –2 / (x – 5), and determine whether the inverse is also a function.
Since the variable is in the denominator, this
is a rational function. Here's the algebra:
the denominator up to the left-hand side of the equation:
the y through the parentheses:
I get the
x-stuff by itself on one side of the "equals" sign:
solve for x:
switch the x's and y's:
This is just another rational function.
The inverse function is y = (5x – 2) / x
Find the inverse of f(x) = –sqrt(x – 2), x> 2. Determine whether
the inverse is also a function, and find the domain and range of the inverse.
The domain restriction
comes from the fact that x
is inside a square root. Usually I wouldn't
bother writing down "x> 2", because I know
that x-values less than 2 would give
me negatives inside the square root. But the restriction is useful in this case because,
together with the graph, it will help me determine the domain and range on the inverse:
The domain is x> 2;
the range (from the graph) is y< 0. Then the domain of the inverse
will be x<
0; the range will be y> 2.
Here's the algebra:
"f(x)" as "y":
x and y:
"y" as "f-inverse".
Since I already figured out the domain and range, I know which half of the quadratic
I have to choose:
inverse y = x2
+ 2 is a function, with domain x<
0 and range y> 2.