Question 1:
Find the parameters a and b included in the linear function f(x) = a x + b so that f^{ 1} (2) = 3 and f^{ 1} (3) = 6, where f^{ 1} (x) is the inverse of function f.
Solution to Question 1:

From the properties of inverse functions if f^{ 1} (2) = 3 and f^{ 1} (3) = 6, then
f(3) = 2 and f(6) =  3

Use the above to write
f(3) = 3a + b = 2 and f(6) = 6a + b = 3

Solve the 2 by 2 system of equations 3a + b = 2 and 6a + b = 3 to obtain
a =  5 / 3 and b = 7
Question 2:
Given f(x) = x^{3} + 2 x, complete the table of values given below and find f ^{1}(3) and f ^{1}( 12).
Solution to Question 2:

Since f(1) = 3, from the properties of the inverse functions, we have
f ^{1}(3) = 1

f is odd function and therefore f( 2) =  f(2) = 12. Hence
f ^{1}( 12) =  2
Question 3: Prove that the inverse of an invertible odd function is also an odd function.
Solution to Question 3:

Start with the property of f and its inverse f^{ 1}
f ( f^{ 1}(x)) = x

Change the right side x of the above equation to  (x) and write
f ( f^{ 1}(x)) =  (  x)

Again change  x in the above equation to f ( f^{ 1}(  x)) and write
f ( f^{ 1}(x)) =  f ( f^{ 1}(  x))

Since f is odd, the right side in the above equation may written as follows
 f ( f^{ 1}(  x)) = f(  f^{ 1}(  x) )

Hence
f ( f^{ 1}(x)) = f(  f^{ 1}(  x) )

Which gives
f^{ 1}(x) =  f^{ 1}(  x)

and proves that f^{ 1} is also odd.
Question 4:
Let f(x) = 1 / (x  2). Find the points of intersection of the graphs of f and that of f^{ 1} the inverse of function f. Graph f, its inverse and the line y = x. Where are the points of intersection located?
Solution to Question 4:

We first find the formula for f^{ 1}(x)
y = 1 / (x  2)

Change y into x and x into y.
x = 1 / (y  2)

Solve the above for y.
y = 1 / x + 2 = f^{ 1}(x)

To find the points of intersection of the graphs of f and f^{ 1}, we need to solve for x the equation
f(x) = f^{ 1}(x)
1 / (x  2) = 1 / x + 2

The above equation has the solutions
x = 1 + √2 and x = 1  + √2

The y coordinate is given by
for x = 1 + √2 , y = f(1 + √2) = 1 + √2
for x = 1  √2 , y = f(1  √2) = 1  √2

The points of intersections are given by their x and y coordinates as follows
(1 + √2 , 1 + √2) and (1  √2 , 1  √2)
The graph of f (in red) and that of f^{ 1} (in blue) are plotted below, along with y = x and the points of intersection which are located on the line y = x.
Question 5:
Graph function f defined by f(x) = x  2 + 2x and its inverse f ^{1} and find a formula for its inverse.
Solution to Question 5:

For (x  2) < 0, x  2 =  (x  2) and f(x) is given by(x)
f(x) =  (x  2) + 2x = x + 2

For x  2 ≥ 0, x  2 = (x  2) f(x) is given by(x)
f(x) = x  2 + 2x = 3x  2

The graph of f (in black) is made up of two linear parts. In order to graph the inverse of , we need to determine points (a,b) on the graph of and then use them to graph the inverse. The points are
(2 , 0) , (2 , 4) , (3 , 7)

On the graph of the inverse function, these points become
(0 , 2) , (4 , 2) , (7 , 3)

Graph them and complete the graph of the inverse function (in blue) using reflection on the line y = x. The graph of the inverse is also made up of two linear parts as shown in the figure below.

If we examine the formula of f and its graph we can assume that the inverse function has a formula of the form:
f ^{1}(x) = ax  4 + bx + c

Coefficient a, b and c are determined using the points (0 , 2) , (4 , 2) , (7 , 3) on the graph of f ^{1} as follows:
f ^{1}(0) = a0  4 + b(0) + c =  2
f ^{1}(4) = a4  4 + b(4) + c = 2
f ^{1}(x) = a7  4 + b(7) + c = 3

We now solve the system of the above equations with unknown a, b and c to find the values:
a =  1 / 3 , b = 2 / 3, c =  2 / 3

Which gives f ^{1}(x) = (1/3)x  4 + 2 x / 3  2 / 3
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