Show \(f(x)=9x^2+6x-5\) is Invertible on \(\mathbb{N}\to S\) and Find \(f^{-1}\)
📺 Video Explanation
📝 Question
Let:
\[ f:\mathbb{N}\to\mathbb{N},\qquad f(x)=9x^2+6x-5 \]
Let \(S\) be the range of \(f\). Show that:
\[ f:\mathbb{N}\to S \]
is invertible. Find:
\[ f^{-1} \]
Also find:
\[ f^{-1}(43),\qquad f^{-1}(163) \]
✅ Solution
🔹 Step 1: Show that \(f:\mathbb{N}\to S\) is one-one
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:
\[ 9(x_1+x_2)+6>0 \]
Therefore:
\[ x_1=x_2 \]
So \(f\) is one-one.
🔹 Step 2: Show that \(f:\mathbb{N}\to S\) is onto
By definition:
\[ S=\text{Range}(f) \]
Every element of \(S\) has a pre-image in \(\mathbb{N}\).
Therefore:
\[ f:\mathbb{N}\to S \]
is onto.
Hence \(f\) is bijective and invertible.
🔹 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:
\[ 3x+1=\sqrt{y+6} \]
So:
\[ x=\frac{\sqrt{y+6}-1}{3} \]
Therefore:
\[ \boxed{f^{-1}(y)=\frac{\sqrt{y+6}-1}{3}} \]
🔹 Step 4: Find required values
For:
\[ y=43 \]
\[ f^{-1}(43)=\frac{\sqrt{43+6}-1}{3} \]
\[ =\frac{\sqrt{49}-1}{3} \]
\[ =\frac{7-1}{3}=2 \]
\[ \boxed{f^{-1}(43)=2} \]
For:
\[ y=163 \]
\[ f^{-1}(163)=\frac{\sqrt{163+6}-1}{3} \]
\[ =\frac{\sqrt{169}-1}{3} \]
\[ =\frac{13-1}{3}=4 \]
\[ \boxed{f^{-1}(163)=4} \]
🎯 Final Answer
\[ \boxed{f^{-1}(y)=\frac{\sqrt{y+6}-1}{3}} \]
and:
\[ \boxed{f^{-1}(43)=2,\qquad f^{-1}(163)=4} \]
Hence \(f:\mathbb{N}\to S\) is invertible.
🚀 Exam Shortcut
- Range as codomain makes function onto automatically
- Complete square to get inverse
- Substitute values directly