Show \(f(x)=5x^2+6x-9\) is Invertible on \(\mathbb{R}_+\) and Find \(f^{-1}\)
📺 Video Explanation
📝 Question
Let:
\[ f:\mathbb{R}_+\to[-9,\infty),\qquad f(x)=5x^2+6x-9 \]
where \(\mathbb{R}_+\) denotes the set of all non-negative real numbers.
Prove that \(f\) is invertible and:
\[ f^{-1}(y)=\frac{\sqrt{54+5y}-3}{5} \]
✅ Solution
🔹 Step 1: Show that \(f\) is one-one
Take:
\[ x_1,x_2\in\mathbb{R}_+ \]
Assume:
\[ f(x_1)=f(x_2) \]
Then:
\[ 5x_1^2+6x_1-9=5x_2^2+6x_2-9 \]
So:
\[ 5(x_1^2-x_2^2)+6(x_1-x_2)=0 \]
Factor:
\[ (x_1-x_2)\big[5(x_1+x_2)+6\big]=0 \]
Since:
\[ x_1,x_2\ge0 \]
we have:
\[ 5(x_1+x_2)+6>0 \]
Therefore:
\[ x_1=x_2 \]
Hence \(f\) is one-one.
🔹 Step 2: Show that \(f\) is onto
Let:
\[ y\in[-9,\infty) \]
Take:
\[ x=\frac{\sqrt{54+5y}-3}{5} \]
Since:
\[ y\ge-9 \Rightarrow 54+5y\ge9 \]
So:
\[ \sqrt{54+5y}\ge3 \]
Hence:
\[ x\ge0 \]
Thus:
\[ x\in\mathbb{R}_+ \]
Now:
\[ 5x+3=\sqrt{54+5y} \]
Square:
\[ 25x^2+30x+9=54+5y \]
So:
\[ 25x^2+30x-45=5y \]
Divide by 5:
\[ 5x^2+6x-9=y \]
Thus:
\[ f(x)=y \]
Hence \(f\) is onto.
🔹 Step 3: Find inverse
Let:
\[ y=5x^2+6x-9 \]
Multiply by 5:
\[ 5y=25x^2+30x-45 \]
Add 54:
\[ 54+5y=25x^2+30x+9 \]
\[ 54+5y=(5x+3)^2 \]
Take square root:
\[ 5x+3=\sqrt{54+5y} \]
Since \(x\ge0\), positive root is taken:
\[ x=\frac{\sqrt{54+5y}-3}{5} \]
Therefore:
\[ \boxed{f^{-1}(y)=\frac{\sqrt{54+5y}-3}{5}} \]
🎯 Final Answer
\[ \boxed{f^{-1}(y)=\frac{\sqrt{54+5y}-3}{5}} \]
Hence \(f\) is invertible.
🚀 Exam Shortcut
- Factor injective proof
- Complete square for inverse
- Use positive root due to domain restriction