Questions: If f(x)=√(x-1) and g(x)=1/(x^2+1) then g[f(x)]= a) 1/x b) 1/√(x-1+1) c) 1/(x-1) d) -1/√(x^2+1)

If f(x)=√(x-1) and g(x)=1/(x^2+1) then g[f(x)]= a) 1/x b) 1/√(x-1+1) c) 1/(x-1) d) -1/√(x^2+1)
Transcript text: If $f(x)=\sqrt{x-1}$ and $g(x)=\frac{1}{x^{2}+1}$ then $g[f(x)]=$ a) $\frac{1}{x}$ b) $\frac{1}{\sqrt{x-1+1}}$ c) $\frac{1}{x-1}$ d) $\frac{-1}{\sqrt{x^{2}+1}}$
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Solution

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Solution Steps

To find g[f(x)] g[f(x)] , we need to substitute f(x) f(x) into g(x) g(x) . Given f(x)=x−1 f(x) = \sqrt{x-1} and g(x)=1x2+1 g(x) = \frac{1}{x^2 + 1} , we substitute x−1 \sqrt{x-1} into g(x) g(x) to get g(x−1) g(\sqrt{x-1}) .

Step 1: Define the Functions

Given the functions: f(x)=x−1 f(x) = \sqrt{x - 1} g(x)=1x2+1 g(x) = \frac{1}{x^2 + 1}

Step 2: Substitute f(x) f(x) into g(x) g(x)

We need to find g(f(x)) g(f(x)) . Substitute f(x)=x−1 f(x) = \sqrt{x - 1} into g(x) g(x) : g(f(x))=g(x−1)=1(x−1)2+1 g(f(x)) = g(\sqrt{x - 1}) = \frac{1}{(\sqrt{x - 1})^2 + 1}

Step 3: Simplify the Expression

Simplify the expression: g(x−1)=1x−1+1=1x g(\sqrt{x - 1}) = \frac{1}{x - 1 + 1} = \frac{1}{x}

Final Answer

The answer is 1x\boxed{\frac{1}{x}}, which corresponds to option (a).

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