Prove \(f \circ f=f\) for \(f(x)=|x|\)

📺 Video Explanation

📝 Question

If:

\[ f(x)=|x| \]

prove that:

\[ f\circ f=f \]

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✅ Solution

🔹 Step 1: Find \((f\circ f)(x)\)

By definition:

\[ (f\circ f)(x)=f(f(x)) \]

Since:

\[ f(x)=|x| \]

Substitute:

\[ (f\circ f)(x)=f(|x|) \]

Again using \(f(t)=|t|\):

\[ (f\circ f)(x)=||x|| \]

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🔹 Step 2: Use property of absolute value

The absolute value of any real number is always non-negative. Applying absolute value again does not change it:

\[ ||x||=|x| \]

for every real number \(x\). This is the idempotent property of the absolute value function. :contentReference[oaicite:1]{index=1}

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🎯 Final Answer

\[ (f\circ f)(x)=||x||=|x|=f(x) \]

Therefore:

\[ \boxed{f\circ f=f} \]

Hence, the absolute value function is idempotent. :contentReference[oaicite:2]{index=2}

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🚀 Exam Shortcut

• First apply modulus: result becomes non-negative
• Second modulus does nothing
• So \(||x||=|x|\)