Questions with Solutions
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) = x3 + 2 x, complete the table of values given below and find f -1(3) and f -1(- 12).|
x |
f(x) |
| 0 | 0 |
| 1 | 3 |
| 2 | 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) = a|x - 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) = a|0 - 4| + b(0) + c = - 2
f -1(4) = a|4 - 4| + b(4) + c = 2
f -1(x) = a|7 - 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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