The function and its inverse are reflections in y=x.
A
a
Domain: x≥ 0.5 Range: y≥ 3
B
b
g(x)=(x-3)^2+1/2
C
c
Domain: x≥ 0.5 Range: y≥ 3
D
d
f(g(x))=x
E
e
g(f(x))=x Explanation: See solution.
Practice makes perfect
a
To graph the function we will start by making a table of values. Since we cannot take the square root of a negative number, x must be greater than or equal to 12. Any value less than that would produce a negative argument under the square root.
x
3+sqrt(2x-1)
f(x)
0.5
3+sqrt(2( 0.5)-1)
3
1
3+sqrt(2( 1)-1)
4
5
3+sqrt(2( 5)-1)
6
13
3+sqrt(2( 13)-1)
8
Let's mark these points in a coordinate plane and draw our graph.
Examining the graph, we notice that the domain is all values greater than or equal to 0.5. Additionally, the range is all values greater than or equal to 3.
Domain f(x): & x ≥ 0.5
Range f(x): & y ≥ 3
Every point that falls on f(x) has a corresponding point on its inverse where the x- and y-coordinates of the points on f(x) have switched position. Using the table of values from Part A, we can identify four points on the inverse.
f(x)
g(x)
(0.5,3)
(3,0.5)
(1,4)
(4,1)
(5,6)
(6,5)
(13,8)
(8,13)
Just like we must switch position of the x- and y-coordinate of the points on f(x) to get some points on g(x), we must also switch the domain and range. In other words, the domain for the function becomes the range of the inverse, and vice versa.
Domain g(x): & x ≥ 3
Range g(x): & y ≥ 0.5
Now we can graph the inverse.
d
Here we have a composite function. The inverse has been substituted into the function.
f( g(x))=3+sqrt(2( (x-3)^2+1/2)-1)
Let's simplify the right-hand side.
