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

Step 1: Define the Functions

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

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

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

Step 3: Simplify the Expression

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

Final Answer

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

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