
Below is shown the graph of f(x) = 2 x^{ 3}  1
.
1) Sketch the graph of the inverse of f in the same system of axes.
2) Find the inverse of and check your answer using some points.
Solution
1) Locate few points on the graph of f. Here is a list of points whose coordinates (a , b) can easily be determined from the graph:
(1 , 1) , (0 , 1) , (1 , 3)
On the graph of the inverse function, the above points will have coordinates (b , a) as follows:
(1 , 1) , (1 , 0) , (3 , 1)
Plot the above points and sketch the graph of the inverse of f so that the two graphs are reflection of each other on the line y = x as shown below.
.
2) Write the given function f(x) = 2 x^{ 3}  1 as an equation in two unknowns.
y = 2 x^{ 3}  1
Solve the above for x.
2 x^{ 3} = y + 1
x^{ 3} = (y + 1) / 2
$$x = \sqrt[3]{\dfrac{y + 1}{2}} $$
Interchange x and y and write the equation of inverse function f^{ 1}:
\( y = \sqrt[3]{\dfrac{x + 1}{2}} \)
\( f^{1}(x) = \sqrt[3]{\dfrac{x + 1}{2}} \)
We now verify that the points (1 , 1) , (1 , 0) and (3 , 1) used above to sketch the graph of the inverse function are on the graph of f^{ 1}.
\( f^{1}(1) = \sqrt[3]{\dfrac{1 + 1}{2}} = 1\)
\( f^{1}(1) = \sqrt[3]{\dfrac{1 + 1}{2}} = 0\)
\( f^{1}(3) = \sqrt[3]{\dfrac{3 + 1}{2}} = 1\)

Let f(x) = x^{ 2}  4 x + 5, x ≤ 2.
1) Find the inverse function of f.
2) Find the domain and the range of f^{ 1}.
Solution
1) We are given a quadratic function with a restricted domain. We first write the given function in vertex form (may be done by completing the square):
f(x) = x^{ 2}  4 x + 5 = (x  2)^{ 2} + 1 , x ≤ 2
The graph of function f is that of the left half of a parabola with vertex at (2 , 1) as shown below.
.
We now write the given function as an equation.
y =(x  2)^{ 2} + 1
Solve the above for x.
y =(x  2)^{ 2} + 1
(x  2)^{ 2} = y  1
Two solutions for x  2: x  2 = +√(y  1) or x  2 =  √(y  1)
x = √(y  1) + 2 or x =  √(y  1) + 2
Since x ≤ 2 (domain of f), we select the solution
x =  √(y  1) + 2
Interchange x and y to write the inverse of function f as follows.
y = f^{ 1}(x) =  √(x  1) + 2
The domain and range of f^{ 1} are the range and domain of f.
Domain of f^{ 1} is the range of f: [1 , +∞) (from graph)
Range of f^{ 1} is the domain of f: (∞ , 2] (given)

Below is shown the graph of f(x) = √(2 x  3).
.
1) Sketch the inverse of f in the same graph.
2) Find the inverse of and check your answer using some points.
Solution
1) Locate few points on the graph of f. A possible list of points whose coordinates (a , b) is as follows:
(1.5 , 0) , (2 , 1) , (6 , 3)
On the graph of the inverse function, the above points will have coordinates (b , a) as follows:
(0 , 1.5) , (1 , 2) , (3 , 6)
Plot the above points and sketch the graph of the inverse of f so that the two graphs are reflection of each other on the line y = x as shown below.
.
2) Write the given function f(x) = √(2 x  3) as an equation in two unknowns.
y = √(2 x  3)
Solve the above for x. First square both sides
2 x  3 = y^{ 2}
2 x = y^{ 2} + 3
x = (y^{ 2} + 3) / 2
Interchange x and y and write the equation of the inverse function f^{ 1}; and write the domain of the inverse.
y = (x^{ 2} + 3) / 2
f^{ 1} (x) = (x^{ 2} + 3) / 2 , x ≥ 0 (domain which is the range of f from its graph above)
We now verify that the points (0 , 1.5) , (1 , 2) and (3 , 6) used to sketch the graph of the inverse function are on the graph of f^{ 1}.
f^{ 1}(0) = (0^{ 2} + 3) / 2 = 1.5
f^{ 1}(1) = (1^{ 2} + 3) / 2 = 2
f^{ 1}(3) = (3^{ 2} + 3) / 2 = 6

Sketch the graph of f^{ 1} using the graph of y = f(x) shown below and find f^{ 1}(x).
.
Solution
1) Use the graph to find points on the graph of f. A possible list of points whose coordinates (a , b) is as follows:
(0 , 3) , (2 , 1) , (5 ,  3)
On the graph of the inverse function, the above points will have coordinates (b , a) as follows:
(3 , 0) , ( 1 , 2) , ( 3 , 5)
Plot the above points and sketch the graph of the inverse of f so that the two graphs are reflection of each other on the line y = x as shown below.
.
2) We now determine f^{ 1}(x). For 3 ≤ x ≤  1 , f^{ 1}(x) has a linear expression with slope m_{1} through the points ( 1 , 2) , ( 3 , 5) given by
m_{1} = (5  2) / (3  (1)) =  3 / 2
For 3 ≤ x ≤  1, f^{ 1}(x) is given by:
f^{ 1}(x) =  (3 / 2)(x  (1)) + 2 =  (3 / 2)(x + 1) + 2
For  1 < x ≤ 3 , f^{ 1}(x) has a linear expression with slope through the points ( 1 , 2) , (3 , 0) given by
m_{2} = (0  2) / (3  (1)) =  1 / 2
For  1 < x ≤ 3, f^{ 1}(x) is given by:
f^{ 1}(x) =  (1 / 2)(x  (1)) + 2 =  (1 / 2)(x + 1) + 2

The one to one function $$f(x) = \sqrt{\dfrac{2}{x}1} $$ is graphed below.
.
1) What is the domain and range of f?
2) Sketch the graph of f^{ 1}.
3) Find f^{ 1}(x) (include domain).
Solution
1) f(x) is defined as a real number if the radicand 2 / x  1 is greater than or equal to 0. Hence we need to solve the inequality:
2 / x  1 ≥ 0
(2  x) / x ≥ 0
The expression on the left of the inequality changes sign at the zeros of the numerator and denominator which are x = 2 and x = 0. See table below.
.
Domain: (0 , 2]
Range: (∞ , 0]
2) Points on the graph of f
(2 , 0) , (1 , 1)
The above points on the graph of the inverse function, will have coordinates (b , a) as follows:
(0 , 2) , ( 1 , 1)
Plot the above points and sketch the graph of the inverse of f so that the two graphs are reflection of each other on the line y = x as shown below.
.
3) Write f(x) as an equation in y and x.
\( y = \sqrt{\dfrac{2}{x}1} \)
Solve the above equation for x. Square both sides of the above equation
\( y^2 = \dfrac{2}{x}1 \)
\( \dfrac{2}{x} = y^2 + 1 \)
\( x = \dfrac{2}{y^2 + 1} \)
Interchange x and y and write the inverse function
\( y = \dfrac{2}{x^2 + 1} \)
\( f^{1}(x) = \dfrac{2}{x^2 + 1} \)
Domain and range of f^{1} are the range and domain of f . Hence
Domain of f^{ 1}: (∞ , 0]
Range of f^{ 1}: (0 , 2]

Below are shown the graph of 6 functions. Sketch the graph of the inverse of each function.
.
Solution
For each graph, select points whose coordinates are easy to determine. Use these points and also the reflection of the graph of function f and its inverse on the line y = x to skectch to sketch the inverse functions as shown below
.

Find the inverse of f(x) = Log_{4}(x + 2)  5, its domain and range.
Solution
Write the given function as an equation in x and y as follows:
y = Log_{4}(x + 2)  5
Solve the above equation for x.
Log_{4}(x + 2) = y + 5
x + 2 = 4^{ (y + 5)}
x = 4^{ (y + 5)}  2
Interchange x and y.
y = 4^{ (x + 5)}  2
Write the inverse function with its domain and range.
f^{1}(x) = 4^{ (x + 5)}  2 , Domain: (∞ , +∞) , Range: (2 , +∞)

If f(x) = ln(x) + 4 x  8, what is the value of f^{ 1}( 4)?
Solution
Let a = f^{ 1}( 4). Then
f(a) = f(f^{ 1}( 4)) =  4 (Using the property f(f^{ 1}(x)) = x of the inverse function).
We now need to find a such that f(a) =  4 hence the equation to solve.
ln(a) + 4 a  8 =  4
ln(a) = 4  4 a
The above equation cannot be solved analytically but its solution may be approximated graphically as the x coordinate of the point of intersection of the graphs of y = ln(x) and y = 4  4x as shown below.
.
The intersection of the two graphs is close to x = 1 which can easily be checked that it is the exact solution to the equation ln(x) = 4  4 x. Hence
f^{1}(  4) = 1

