Given that $H(x) = f(\sqrt{x})$, this equation defines a composite function, where the input to the function $f$ is the square root of $x$. The function $H(x)$ represents a transformation of $f(x)$, where $x$ is replaced by $\sqrt{x}$.

Understanding the Composition:

1. Domain of $H(x)$:
   Since $\sqrt{x}$ is only defined for $x \geq 0$, the domain of $H(x)$ will also be limited to non-negative values of $x$. Hence, the domain of $H(x)$ is $[0, \infty)$.

2. Behavior of $H(x)$:

   • When $x = 0$, $H(x) = f(0)$.
   • For $x > 0$, $H(x)$ depends on how $f$ behaves at $\sqrt{x}$. This means that $H(x)$ transforms the behavior of $f$, compressing or expanding it according to the square root function.

Would you like to know how this function behaves for a specific $f(x)$? Or any other details?

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Here are some related questions:

1. What is the derivative of $H(x)$ in terms of $f'(x)$?
2. How would the graph of $H(x)$ look for different $f(x)$?
3. How does the domain and range of $H(x)$ compare to $f(x)$?
4. If $f(x)$ is an exponential function, what would be the effect on $H(x)$?
5. Can we express an inverse function for $H(x)$?

Tip: When dealing with composite functions, always check how the inner function affects the domain and range of the outer function!