To find the expression for \(\frac{f}{g}(x)\), where \(f(x) = x^2 + \sqrt{x}\) and \(g(x) = \frac{x^2}{x-1}\), we need to divide the function \(f(x)\) by the function \(g(x)\).

The expression for \(\frac{f}{g}(x)\) is given by:

\[
\frac{f}{g}(x) = \frac{f(x)}{g(x)} = \frac{x^2 + \sqrt{x}}{\frac{x^2}{x-1}}
\]

To simplify, multiply both the numerator and the denominator by \(x-1\):

\[
\frac{f}{g}(x) = \frac{(x^2 + \sqrt{x}) \cdot (x-1)}{x^2}
\]

Expanding the numerator:

\[
(x^2 + \sqrt{x}) \cdot (x-1) = x^3 - x^2 + x^{\frac{3}{2}} - \sqrt{x}
\]

So the final simplified expression for \(\frac{f}{g}(x)\) is:

\[
\frac{f}{g}(x) = \frac{x^3 - x^2 + x^{\frac{3}{2}} - \sqrt{x}}{x^2}
\]

This can further be simplified by dividing each term in the numerator by \(x^2\):

\[
\frac{f}{g}(x) = x - 1 + \frac{x^{\frac{1}{2}}}{x} - \frac{\sqrt{x}}{x^2}
\]

Finally:

\[
\frac{f}{g}(x) = x - 1 + \frac{1}{\sqrt{x}} - \frac{1}{x^{\frac{3}{2}}}
\]

Would you like further details or have any questions?

### 5 Related Questions:
1. How do you find the domain of the function \(\frac{f}{g}(x)\)?
2. What are the possible points of discontinuity for \(\frac{f}{g}(x)\)?
3. How can you find the derivative of \(\frac{f}{g}(x)\)?
4. What is the behavior of \(\frac{f}{g}(x)\) as \(x\) approaches infinity?
5. Can \(\frac{f}{g}(x)\) be simplified further if \(x\) has specific values?

### Tip:
When simplifying expressions involving square roots, remember to express all terms in a consistent form, such as using exponents to combine and simplify terms.

Explore how to find the composition of functions f(x) = x^2 + sqrt(x) and g(x) = x^2 / (x - 1). Learn step-by-step the simplification process of f/g(x). Suitable for advanced high school students and beyond.

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