Questions and Solutions
Question 1
Find the parameters \(a\) and \(b\) of the linear function \[ f(x) = ax + b \] such that \[ f^{-1}(2) = 3 \quad \text{and} \quad f^{-1}(-3) = 6. \]
View Solution
- From the definition of inverse functions: \[ f^{-1}(2) = 3 \Rightarrow f(3) = 2, \quad f^{-1}(-3) = 6 \Rightarrow f(6) = -3. \]
- Substitute into \(f(x) = ax + b\): \[ 3a + b = 2, \quad 6a + b = -3. \]
- Solving the system gives: \[ a = -\frac{5}{3}, \quad b = 7. \]
Question 2
Given that \( f(x) \) is an odd function and
| \(x\) | \(f(x)\) |
|---|---|
| 0 | 0 |
| 1 | 3 |
| 2 | 12 |
Find \(f^{-1}(3)\) and \(f^{-1}(-12)\).
View Solution
- Since \(f(1) = 3\), it follows that: \[ f^{-1}(3) = 1. \]
- The function \(f\) is odd, so: \[ f(-2) = -f(2) = -12 \Rightarrow f^{-1}(-12) = -2. \]
Question 3
Prove that the inverse of an invertible odd function is also odd.
View Solution
- By definition of inverse functions: \[ f(f^{-1}(x)) = x. \]
- Replace \(x\) with \(-x\): \[ f(f^{-1}(-x)) = -x. \]
- Since \(f\) is odd: \[ f(-u) = -f(u), \] which implies: \[ f(-f^{-1}(-x)) = x. \]
- Comparing both expressions: \[ f^{-1}(x) = -f^{-1}(-x), \] proving that \(f^{-1}\) is odd.
Question 4
Let \[ f(x) = \frac{1}{x - 2}. \] Find the points of intersection of the graphs of \(f\) and its inverse. Graph \(f\), \(f^{-1}\), and the line \(y = x\).
View Solution
- Start with: \[ y = \frac{1}{x - 2}. \]
- Swap \(x\) and \(y\): \[ x = \frac{1}{y - 2}. \]
- Solve for \(y\): \[ y = \frac{1}{x} + 2 = f^{-1}(x). \]
- Solve: \[ \frac{1}{x - 2} = \frac{1}{x} + 2. \] This gives: \[ x = 1 \pm \sqrt{2}. \]
- The intersection points are: \[ (1 + \sqrt{2},\, 1 + \sqrt{2}), \quad (1 - \sqrt{2},\, 1 - \sqrt{2}). \]
Question 5
Graph the function \[ f(x) = |x - 2| + 2x, \] find its inverse, and graph both.
View Solution
- For \(x < 2\): \[ f(x) = -(x - 2) + 2x = x + 2. \]
- For \(x \ge 2\): \[ f(x) = (x - 2) + 2x = 3x - 2. \]
- From the graph, sample points on \(f\) are: \[ (-2,0),\ (2,4),\ (3,7). \]
- Corresponding points on \(f^{-1}\) are: \[ (0,-2),\ (4,2),\ (7,3). \]
- Solving for the inverse gives: \[ f^{-1}(x) = -\frac{1}{3}|x - 4| + \frac{2}{3}x - \frac{2}{3}. \]
