Suppose $$ H(x)=5 \sqrt{x}-3 $$, Find two functions $$ f $$ and $$ g $$ such that $$ (f \circ g)(x)=H(x) $$. Neither function can be the identity function. (There may be more than one correct answer.) $$ f(x)=\square $$ $$ g(x)=\square $$

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We are given 
$$
H(x)=5\sqrt{x}-3.
$$
We want to find functions $$ f $$ and $$ g $$ such that
$$
(f\circ g)(x)=f\bigl(g(x)\bigr)=H(x)=5\sqrt{x}-3.
$$
One possibility is to let
$$
g(x)=\sqrt{x}.
$$
Then choose
$$
f(x)=5x-3.
$$
Let's verify the composition:
$$
(f\circ g)(x)=f(g(x))=5\left(\sqrt{x}\right)-3=5\sqrt{x}-3,
$$
which matches $$ H(x) $$.

Thus, one valid answer is:
$$
\text{$$ f(x)=5x-3 $$} \quad \text{and} \quad \text{$$ g(x)=\sqrt{x} $$}.
$$
Both functions are non-identity, satisfying the given requirements.

Another possibility (for example) is to choose
$$
g(x)=5\sqrt{x} \quad \text{and} \quad f(x)=x-3.
$$
Here, 
$$
(f\circ g)(x)=f(5\sqrt{x})=5\sqrt{x}-3,
$$
which again matches $$ H(x) $$.

Both answers are correct, but one valid answer is:
$$
f(x)=5x-3,\qquad g(x)=\sqrt{x}.
$$
One possible solution is: $$ f(x) = 5x - 3 $$ and $$ g(x) = \sqrt{x} $$.

Suppose ,
Find two functions and such that .
Neither function can be the identity function.
(There may be more than one correct answer.)

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Suppose , Find two functions and such that . Neither function can be the identity function. (There may be more than one correct answer.)