Solution;
One possible way ;
If you let;
f ( x ) = x f(x)=\sqrt{x} f ( x ) = x
g ( x ) = x 2 + x g(x)=\sqrt{x^2+x} g ( x ) = x 2 + x
h ( x ) = 1 x h(x)=\frac1x h ( x ) = x 1
First,obtain g of f(x);
g o f = g ( f ( x ) ) = ( f ( x ) ) 2 + f ( x ) = ( x ) 2 + x = x + x g o f=g(f(x))=\sqrt{(f(x))^2+f(x)}=\sqrt{(\sqrt{x})^2+\sqrt{x}}=\sqrt{x+\sqrt{x}} g o f = g ( f ( x )) = ( f ( x ) ) 2 + f ( x ) = ( x ) 2 + x = x + x
Now we plug g(f(x)) into h(x);
F ( x ) = h ( g o f ) = h ( g ( f ( x ) ) ) F(x)=h (gof)=h(g(f(x))) F ( x ) = h ( g o f ) = h ( g ( f ( x )))
F ( x ) = 1 g ( f ( x ) ) F(x)=\frac{1}{g(f(x))} F ( x ) = g ( f ( x )) 1 =1 x + x \frac{1}{\sqrt{x+\sqrt{x}}} x + x 1
Hence the expression of the function is three functions could be expressed as;
f ( x ) = x f(x)=\sqrt{x} f ( x ) = x ,g ( x ) = x 2 + x g(x)=\sqrt{x^2+x} g ( x ) = x 2 + x , h ( x ) = 1 x h(x)=\frac1x h ( x ) = x 1 ,F ( x ) = h F(x)=h F ( x ) = h o g g g o f f f
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