Show \(f(x)=x^3-3\) is Invertible and Find \(f^{-1}\)
📺 Video Explanation
📝 Question
Let:
\[ f:\mathbb{R}\to\mathbb{R},\qquad f(x)=x^3-3 \]
Prove that inverse exists and find:
\[ f^{-1}(x) \]
Also find:
\[ f^{-1}(24),\qquad f^{-1}(5) \]
✅ Solution
🔹 Step 1: Show that \(f\) is one-one
Assume:
\[ f(x_1)=f(x_2) \]
Then:
\[ x_1^3-3=x_2^3-3 \]
So:
\[ x_1^3=x_2^3 \]
Taking cube root:
\[ x_1=x_2 \]
Therefore:
\[ f \text{ is one-one} \]
🔹 Step 2: Show that \(f\) is onto
Let:
\[ y\in\mathbb{R} \]
Need:
\[ f(x)=y \]
So:
\[ x^3-3=y \]
\[ x^3=y+3 \]
\[ x=\sqrt[3]{y+3} \]
Since cube root of every real exists:
\[ x\in\mathbb{R} \]
Thus:
\[ f \text{ is onto} \]
🔹 Step 3: Find inverse
Let:
\[ y=x^3-3 \]
Then:
\[ x^3=y+3 \]
So:
\[ x=\sqrt[3]{y+3} \]
Therefore:
\[ \boxed{f^{-1}(x)=\sqrt[3]{x+3}} \]
🔹 Step 4: Evaluate values
\[ f^{-1}(24)=\sqrt[3]{24+3}=\sqrt[3]{27}=3 \]
\[ \boxed{f^{-1}(24)=3} \]
\[ f^{-1}(5)=\sqrt[3]{5+3}=\sqrt[3]{8}=2 \]
\[ \boxed{f^{-1}(5)=2} \]
🎯 Final Answer
\[ \boxed{f^{-1}(x)=\sqrt[3]{x+3}} \]
and:
\[ \boxed{f^{-1}(24)=3,\qquad f^{-1}(5)=2} \]
🚀 Exam Shortcut
- Cubic functions are strictly increasing on \(\mathbb{R}\)
- Solve \(y=x^3-3\) for \(x\)
- Cube root gives inverse