Find \(f^{-1}\) for \(f(x)=x^2\) on Given Finite Sets

📺 Video Explanation

📝 Question

Let:

\[ A=\{1,3,5,7,9\},\qquad B=\{0,1,9,25,49,81\} \]

and:

\[ f:A\to B,\qquad f(x)=x^2 \]

Find:

\[ f^{-1} \]

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✅ Solution

🔹 Step 1: Find image of each element of \(A\)

Using:

\[ f(x)=x^2 \]

We get:

• \(f(1)=1\)
• \(f(3)=9\)
• \(f(5)=25\)
• \(f(7)=49\)
• \(f(9)=81\)

So:

\[ f=\{(1,1),(3,9),(5,25),(7,49),(9,81)\} \]

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🔹 Step 2: Check whether inverse exists

All outputs are distinct.

So:

\[ f \text{ is one-one} \]

But codomain:

\[ B=\{0,1,9,25,49,81\} \]

Range:

\[ \{1,9,25,49,81\} \]

Element:

\[ 0\in B \]

has no pre-image.

Therefore:

\[ f \text{ is not onto} \]

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🎯 Final Answer

Since \(f\) is not onto, it is not bijective.

Therefore:

\[ \boxed{\text{Inverse function does not exist}} \]

(Though inverse relation exists: \[ \{(1,1),(9,3),(25,5),(49,7),(81,9)\} \] but it is not a function from \(B\to A\).)

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🚀 Exam Shortcut

• Inverse exists only if function is bijection
• Check codomain carefully
• Unused codomain element ⇒ no inverse