Show \(f(x)=(x+1)^2-1\) is Invertible and Find Set \(S\)

📝 Question

Let:

\[ f:[-1,\infty)\to[-1,\infty), \quad f(x)=(x+1)^2-1 \]

Show that \(f\) is invertible. Also find the set:

\[ S=\{x: f(x)=f^{-1}(x)\} \]

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✅ Solution

🔹 Step 1: Prove that \(f\) is one-one

The function \(f(x)=(x+1)^2-1\) is increasing on \([-1,\infty)\).

Hence, \(f\) is one-one (injective).

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🔹 Step 2: Range of the function

Since \((x+1)^2 \geq 0\), we have:

\[ f(x)\geq -1 \]

So, range of \(f\) is \([-1,\infty)\).

Thus, \(f\) is onto.

Hence, \(f\) is bijective and invertible. :contentReference[oaicite:0]{index=0}

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🔹 Step 3: Find the inverse

Let:

\[ y=(x+1)^2-1 \]

\[ y+1=(x+1)^2 \]

Taking square root (since \(x\geq -1\)):

:contentReference[oaicite:1]{index=1}

\[ x=\sqrt{y+1}-1 \]

Therefore:

\[ \boxed{f^{-1}(x)=\sqrt{x+1}-1} \]

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🔹 Step 4: Find \(S=\{x: f(x)=f^{-1}(x)\}\)

We solve:

\[ (x+1)^2-1=\sqrt{x+1}-1 \]

Add 1 both sides:

\[ (x+1)^2=\sqrt{x+1} \]

Let \(t=\sqrt{x+1}\) (so \(t\geq 0\)):

\[ t^4=t \]

\[ t(t^3-1)=0 \]

\[ t=0 \quad \text{or} \quad t=1 \]

Now,

If \(t=0\): \(x+1=0 \Rightarrow x=-1\)

If \(t=1\): \(x+1=1 \Rightarrow x=0\)

Thus,

\[ \boxed{S=\{-1,\,0\}} \]

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🎯 Final Answer

\[ \boxed{f^{-1}(x)=\sqrt{x+1}-1} \]

\[ \boxed{S=\{-1,\,0\}} \]

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🚀 Exam Shortcut

• Restrict domain to make quadratic increasing
• Take only positive square root
• Use substitution \(t=\sqrt{x+1}\) to simplify equation
• Solve algebraically and back-substitute