Find \(f \circ g\) and \(g \circ f\) for \(f(x)=\sqrt{1-x}\) and \(g(x)=\ln x\)

📺 Video Explanation

📝 Question

Let functions \(f\) and \(g\) be defined as:

\[ f(x)=\sqrt{1-x},\qquad g(x)=\ln x \]

Find:

• \((f\circ g)(x)\)
• \((g\circ f)(x)\)

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✅ Solution

🔹 Find \((f\circ g)(x)\)

By definition:

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

Substitute \(g(x)=\ln x\):

\[ (f\circ g)(x)=f(\ln x)=\sqrt{1-\ln x} \]

For this to be defined:

• \(\ln x\) must exist, so \(x>0\)
• Inside square root must be non-negative: \[ 1-\ln x\ge 0 \]

So:

\[ \ln x\le 1 \]

\[ x\le e \]

Hence domain:

\[ 0

Therefore:

\[ \boxed{(f\circ g)(x)=\sqrt{1-\ln x},\quad 0

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🔹 Find \((g\circ f)(x)\)

By definition:

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

Substitute \(f(x)=\sqrt{1-x}\):

\[ (g\circ f)(x)=g(\sqrt{1-x})=\ln\left(\sqrt{1-x}\right) \]

For this to be defined:

• Square root must exist: \[ 1-x\ge 0 \Rightarrow x\le 1 \]
• Logarithm argument must be positive: \[ \sqrt{1-x}>0 \]

So:

\[ 1-x>0 \Rightarrow x<1 \]

Hence domain:

\[ x<1 \]

Therefore:

\[ \boxed{(g\circ f)(x)=\ln\left(\sqrt{1-x}\right),\quad x<1} \]

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🎯 Final Answer

\[ \boxed{(f\circ g)(x)=\sqrt{1-\ln x},\quad 0

\[ \boxed{(g\circ f)(x)=\ln\left(\sqrt{1-x}\right),\quad x<1} \]

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🚀 Exam Shortcut

• For square root: inside expression must be \(\ge 0\)
• For logarithm: argument must be \(>0\)
• Check domain after substitution in composite functions