Drawing the Inverse from a Graph
IB Mathematics AA · SL & HL · Shadow Worksheet
Practice
Question 1
Explain why a function must be one-to-one (injective) for its inverse to also be a function. Give an example of a function that is not one-to-one and state the domain restriction needed.
Explanation
If \(f\) is not one-to-one, then two \(x\)-values map to the same \(y\)-value. The inverse would then map one \(y\)-value to two \(x\)-values, violating the definition of a function.
Example
\(f(x)=x^2\): not one-to-one on \(\mathbb{R}\). Restrict to \(x\geq0\) for \(f^{-1}(x)=\sqrt{x}\) to exist.
Question 2
The graph of \(y=f(x)\) passes through \((1,3),(2,5),(3,8),(-1,0)\). Write down four points on the graph of \(y=f^{-1}(x)\).
Answer
\((3,1),(5,2),(8,3),(0,-1)\)
Question 3
For \(f(x)=2x-4\): sketch \(y=f(x)\), \(y=f^{-1}(x)\), and \(y=x\) on the same axes. Label all intercepts.
Find inverse
\(f^{-1}(x)=\dfrac{x+4}{2}\)
\(f(x)\) intercepts
\((2,0)\) and \((0,-4)\)
\(f^{-1}(x)\) intercepts
\((-4,0)\) and \((0,2)\)
Note
Both lines intersect on \(y=x\) at \((4,4)\)
Question 4
Given \(f(x)=e^x-2\), find \(f^{-1}(x)\) and state its domain and range. Sketch both functions.
Find inverse
\(y=e^x-2\Rightarrow x=\ln(y+2)\)
Answer
\(f^{-1}(x)=\ln(x+2)\); Domain: \(x>-2\); Range: \(\mathbb{R}\)
Question 5
Show that \(f(x)=\dfrac{x+1}{x-1}\) is self-inverse (i.e. \(f^{-1}(x)=f(x)\)).
Find \(f^{-1}\)
\(y=\dfrac{x+1}{x-1}\Rightarrow y(x-1)=x+1\Rightarrow x(y-1)=y+1\Rightarrow x=\dfrac{y+1}{y-1}=f(y)\)
Conclusion
\(f^{-1}(x)=\dfrac{x+1}{x-1}=f(x)\) ✓. Self-inverse. ■
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