For $x \in \mathbb{R}$, two real valued functions $f(x)$ and $g(x)$ are such that, $g(x)=\sqrt{x}+1$ and $f \circ g(x)=x+3-\sqrt{x}$. Then $f(0)$ is equal to
Solution
$$
\begin{aligned}
& g(x)=\sqrt{x}+1 \\\\
& \operatorname{fog}(x)=x+3-\sqrt{x} \\\\
& =(\sqrt{x}+1)^2-3(\sqrt{x}+1)+5 \\\\
& =g^2(x)-3 g(x)+5 \\\\
& \Rightarrow f(x)=x^2-3 x+5 \\\\
& \therefore f(0)=5
\end{aligned}
$$
<br/><br/>But, if we consider the domain of the composite function $f \circ g(x)$ then in that case $f(0)$ will be not defined as $\mathrm{g}(\mathrm{x})$ cannot be equal to zero.
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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