Show \(f(x)=x^2+4\) is Invertible on \(\mathbb{R}_+\) and Find \(f^{-1}\)
📺 Video Explanation
📝 Question
Let:
\[ f:\mathbb{R}_+\to[4,\infty),\qquad f(x)=x^2+4 \]
where \(\mathbb{R}_+\) is the set of all non-negative real numbers.
Show that \(f\) is invertible and:
\[ f^{-1}(x)=\sqrt{x-4} \]
Also verify:
\[ f^{-1}\circ f(x)=x \]
✅ Solution
🔹 Step 1: Show that \(f\) is one-one
Let:
\[ f(x_1)=f(x_2) \]
Then:
\[ x_1^2+4=x_2^2+4 \]
So:
\[ x_1^2=x_2^2 \]
Since:
\[ x_1,x_2\in\mathbb{R}_+ \]
both are non-negative. Hence:
\[ x_1=x_2 \]
Therefore:
\[ f \text{ is one-one} \]
🔹 Step 2: Show that \(f\) is onto
Let:
\[ y\in[4,\infty) \]
Then:
\[ y-4\ge0 \]
Take:
\[ x=\sqrt{y-4} \]
Clearly:
\[ x\in\mathbb{R}_+ \]
Now:
\[ f(x)=x^2+4 \]
Substitute:
\[ f(\sqrt{y-4})=(\sqrt{y-4})^2+4=y \]
Hence every element in codomain has pre-image.
So:
\[ f \text{ is onto} \]
🔹 Step 3: Find inverse
Let:
\[ y=x^2+4 \]
Then:
\[ x^2=y-4 \]
Since \(x\ge0\):
\[ x=\sqrt{y-4} \]
Therefore:
\[ \boxed{f^{-1}(x)=\sqrt{x-4}} \]
🔹 Step 4: Verify \(f^{-1}\circ f\)
\[ (f^{-1}\circ f)(x)=f^{-1}(x^2+4) \]
Substitute inverse:
\[ (f^{-1}\circ f)(x)=\sqrt{(x^2+4)-4} \]
\[ =\sqrt{x^2} \]
Since \(x\in\mathbb{R}_+\):
\[ \sqrt{x^2}=x \]
Therefore:
\[ \boxed{f^{-1}\circ f(x)=x} \]
🎯 Final Answer
\[ \boxed{f^{-1}(x)=\sqrt{x-4}} \]
and:
\[ \boxed{(f^{-1}\circ f)(x)=x} \]
Hence \(f\) is invertible.
🚀 Exam Shortcut
- Restrict domain to non-negative values
- Solve \(y=x^2+4\) for \(x\)
- Take positive root only