#### Answer

$h(x)=\sqrt{x}$
$g(x)=1+x$
$f(x)= \frac{1}{\sqrt{x}}$
Other answers are also possible.

#### Work Step by Step

If we let:
$h(x)=\sqrt{x}$
$g(x)=1+x$
$f(x)= \frac{1}{\sqrt{x}}$
Then:
$(f \circ g\circ h)(x)=f(g(h(x))=f(g(\sqrt{x}))=f(1+\sqrt{x})=\frac{1}{\sqrt{1+\sqrt{x}}}=T(x)$
Other answers are also possible.