If $f (x) = \sqrt{x + 3} and g (x) = x^{2} + 1$ be two real functions, then find fog and gof.

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Solution

$f (x) = \sqrt{x + 3} For domain, x + 3 \geq 0 \Rightarrow x \geq - 3 Domain of f =[-3, ∞) Since f is a square root function, range of f =[0, ∞) f : [-3, \infty) \to [0, \infty) g (x) = x^{2} + 1 is a polynomial. \Rightarrow g : R \to R Computation of f o g : Range of g is not a subset of the domain of f . and domain (f o g) = \{x : x \in domain of g and g (x) \in domain of f (x)\} \Rightarrow Domain (f o g) = \{x : x \in R and x^{2} + 1 \in [-3, \infty)\} \Rightarrow Domain (f o g) = \{x : x \in R and x^{2} + 1 \geq - 3\} \Rightarrow Domain (f o g) = \{x : x \in R and x^{2} + 4 \geq 0\} \Rightarrow Domain (f o g) = \{x : x \in R and x \in R\} \Rightarrow Domain (f o g) =R f o g : R \to R (f o g) (x) = f (g (x)) = f (x^{2} + 1) = \sqrt{x^{2} + 1 + 3} = \sqrt{x^{2} + 4} Computation of g o f : Range of f is a subset of the domain of g . g o f : [-3, \infty) \to R \Rightarrow (g o f) (x) = g (f (x)) = g (\sqrt{x + 3}) = {(\sqrt{x + 3})}^{2} + 1 = x + 3 + 1 = x + 4$

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If fx=x+3 and gx=x2+1 be two real functions, then find fog and gof.

If $f (x) = \sqrt{x + 3} and g (x) = x^{2} + 1$ be two real functions, then find fog and gof.