Show \(f(x)=9x^2+6x-5\) is Invertible on \(\mathbb{R}_+\) and Find \(f^{-1}\)
📺 Video Explanation
📝 Question
Let:
\[ f:\mathbb{R}_+\to[-5,\infty),\qquad f(x)=9x^2+6x-5 \]
where \(\mathbb{R}_+\) denotes the set of all non-negative real numbers.
Show that \(f\) is invertible and find:
\[ f^{-1}(x)=\frac{\sqrt{x+6}-1}{3} \]
✅ Solution
🔹 Step 1: Show that \(f\) is one-one
Take:
\[ x_1,x_2\in\mathbb{R}_+ \]
and assume:
\[ f(x_1)=f(x_2) \]
Then:
\[ 9x_1^2+6x_1-5=9x_2^2+6x_2-5 \]
So:
\[ 9(x_1^2-x_2^2)+6(x_1-x_2)=0 \]
Factor:
\[ (x_1-x_2)\big[9(x_1+x_2)+6\big]=0 \]
Since:
\[ x_1,x_2\ge0 \]
we have:
\[ 9(x_1+x_2)+6>0 \]
Therefore:
\[ x_1-x_2=0 \]
Hence:
\[ x_1=x_2 \]
So \(f\) is one-one.
🔹 Step 2: Show that \(f\) is onto
Let:
\[ y\in[-5,\infty) \]
Then:
\[ y+6\ge1>0 \]
Take:
\[ x=\frac{\sqrt{y+6}-1}{3} \]
Since:
\[ y\ge-5 \]
we get:
\[ \sqrt{y+6}\ge1 \]
So:
\[ x\ge0 \]
Hence:
\[ x\in\mathbb{R}_+ \]
Now:
\[ 3x=\sqrt{y+6}-1 \]
So:
\[ 3x+1=\sqrt{y+6} \]
Square:
\[ (3x+1)^2=y+6 \]
Expand:
\[ 9x^2+6x+1=y+6 \]
Thus:
\[ 9x^2+6x-5=y \]
So:
\[ f(x)=y \]
Hence \(f\) is onto.
🔹 Step 3: Find inverse
Let:
\[ y=9x^2+6x-5 \]
Add 6:
\[ y+6=9x^2+6x+1 \]
\[ y+6=(3x+1)^2 \]
Take square root:
\[ \sqrt{y+6}=3x+1 \]
Since \(x\ge0\), choose positive root:
\[ x=\frac{\sqrt{y+6}-1}{3} \]
Therefore:
\[ \boxed{f^{-1}(x)=\frac{\sqrt{x+6}-1}{3}} \]
🎯 Final Answer
\[ \boxed{f^{-1}(x)=\frac{\sqrt{x+6}-1}{3}} \]
Hence \(f\) is invertible on \(\mathbb{R}_+\).
🚀 Exam Shortcut
- Complete square first
- Restrict domain to non-negative values
- Take positive root only