I need to solve for h(x) given that h(x) = (f◦g)-1(x)
f(x) = sqrt(x-8) and g(x) = x + 4
First we find (f◦g)(x) which means the same as f(g(x)).
To get f(g(x)) we start with
f(x) = √x - 8
Then we remove the x's
f( ) = √ - 8
Then we look at g(x) = x+4
We put "g(x)" in place of the x that we removed from the left side of the
equation, and we put "x+4" in place of the x that we removed from the
right side of the equation, and we have this:
f(g(x)) = √(x+4) - 8
Now we simplify the right side:
f◦g(x) = f(g(x)) = √x+4-8
f◦g(x) = √x-4
When we graph f◦g(x), it looks like the red curve below. The green
dotted line is the IDENTITY-line whose equation is y=x (it is the line all
points of which y and x are IDENTICAL. It is the line that the graph will
be reflected into in order to find the graph of the inverse (f◦g)-1(x):
Notice that the domain of f◦g(x) is [4,
) and its
range is [0.). The domain of the inverse (f◦g)-1(x) will be
the same as the range of f◦g(x) and vice-versa.
f◦g(x) = √x-4
Replace f◦g(x) by y
y = √x-4
Interchange x and y
x = √y-4
Solve for y. Begin by squaring both sides to get rid of the
square root:
x² = y-4
Solve for y
x²+4 = y
Turn it around
y = x²+4
Replace y by (f◦g)-1(x)
(f◦g)-1(x) = x²+4 (with restrictions)
You are told to rename (f◦g)-1(x) as h(x)
Now you might think that has domain (
,), as it
ordinarily would have, but it isn't in this case, for its domain must be the
same as the range of f◦g(x), and vice-versa. So we must restrict the
domain to the same as the range of f◦g(x), which is [4,).
So the answer is not just (f◦g)-1(x) = x²+4 , but it is
h(x) = (f◦g)-1(x) = x²+4 where x∈(0,)
You must state the restricted domain.
Now the graph of (f◦g)-1(x) is the reflection in the
IDENTITY line (dotted green line whose equation is y=x:
The domain of the bottom curve f◦g(x) is [4,).
The range of the bottom curve f◦g(x) is [0,).
The domain of the upper curve h(x) = (f◦g)-1(x) is [0,).
The range of the upper curve h(x) = (f◦g)-1(x) is [4,).
Edwin
