Check Whether \(f(x)=x^3+4\) is Bijective and Find \(f^{-1}(3)\)
📺 Video Explanation
📝 Question
Let:
\[ f:\mathbb{R}\to\mathbb{R},\qquad f(x)=x^3+4 \]
Check whether \(f\) is bijection. If yes, find:
\[ f^{-1}(3) \]
✅ Solution
🔹 Step 1: Check one-one
Assume:
\[ f(x_1)=f(x_2) \]
Then:
\[ x_1^3+4=x_2^3+4 \]
So:
\[ x_1^3=x_2^3 \]
Hence:
\[ x_1=x_2 \]
Therefore:
\[ f \text{ is one-one} \]
🔹 Step 2: Check onto
Let:
\[ y\in\mathbb{R} \]
Need:
\[ f(x)=y \]
So:
\[ x^3+4=y \]
\[ x^3=y-4 \]
\[ x=\sqrt[3]{y-4} \]
Since cube root exists for every real number:
\[ x\in\mathbb{R} \]
Hence:
\[ f \text{ is onto} \]
🔹 Step 3: Conclusion
Since \(f\) is both one-one and onto:
\[ f \text{ is bijection} \]
So inverse exists:
\[ f^{-1}(x)=\sqrt[3]{x-4} \]
🔹 Step 4: Find \(f^{-1}(3)\)
\[ f^{-1}(3)=\sqrt[3]{3-4} \]
\[ =\sqrt[3]{-1} \]
\[ \boxed{f^{-1}(3)=-1} \]
🎯 Final Answer
\[ \boxed{f \text{ is bijective}} \]
\[ \boxed{f^{-1}(x)=\sqrt[3]{x-4}} \]
and:
\[ \boxed{f^{-1}(3)=-1} \]
🚀 Exam Shortcut
- Cubic functions are bijections on \(\mathbb{R}\)
- Solve \(y=x^3+4\) for \(x\)
- Use cube root directly