g ( x ) = 1 x + x , g : R + → R + g(x)=\frac{1}{\sqrt{x + \sqrt{x}}}, \quad g: \mathbb{R}^+ \to \mathbb{R}^+ g ( x ) = x + x 1 , g : R + → R +
Let f ( x ) = x , h ( x ) = x 2 + x , and k ( x ) = 1 x . f(x)=\sqrt{x}, h(x)=\sqrt{x^2+x}, \text{ and } k(x) = \frac{1}{x}. f ( x ) = x , h ( x ) = x 2 + x , and k ( x ) = x 1 .
Determine h ∘ f . h\circ f. h ∘ f .
h ∘ f = h ( f ( x ) ) = ( f ( x ) ) 2 + f ( x ) = ( x ) 2 + x = x + x h\circ f = h(f(x)) = \sqrt{(f(x))^2 + f(x)} = \sqrt{(\sqrt{x})^2 + \sqrt{x}} \\
\qquad = \sqrt{x + \sqrt{x}} h ∘ f = h ( f ( x )) = ( f ( x ) ) 2 + f ( x ) = ( x ) 2 + x = x + x
Now, determine k ∘ ( h ∘ f ) . k\circ(h\circ f). k ∘ ( h ∘ f ) .
k ∘ ( h ∘ f ) = k ( h ( f ( x ) ) ) = 1 h ( f ( x ) ) = 1 x + x k\circ(h\circ f) = k(h(f(x))) = \frac{1}{h(f(x))} = \frac{1}{\sqrt{x + \sqrt{x}} } k ∘ ( h ∘ f ) = k ( h ( f ( x ))) = h ( f ( x )) 1 = x + x 1
Therefore, g ( x ) = k ∘ ( h ∘ f ) , where f ( x ) = x , h ( x ) = x 2 + x , and k ( x ) = 1 x . g(x) = k\circ(h\circ f), \text{ where } f(x)=\sqrt{x}, h(x)=\sqrt{x^2+x},
\text{ and } k(x) = \frac{1}{x}. g ( x ) = k ∘ ( h ∘ f ) , where f ( x ) = x , h ( x ) = x 2 + x , and k ( x ) = x 1 .
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