To determine the point(s) at which the function $f(x) = \frac{x^2}{x - 1}$ has a horizontal tangent, we need to follow these steps:

Step 1: Find the derivative $f'(x)$

The horizontal tangent occurs when the derivative of the function is equal to zero, i.e., $f'(x) = 0$.

Given $f(x) = \frac{x^2}{x - 1}$, we'll apply the quotient rule to find the derivative. The quotient rule is:

$\left( \frac{u}{v} \right)' = \frac{v \cdot u' - u \cdot v'}{v^2}$

where $u = x^2$ and $v = x - 1$. Now, let's compute the derivatives:

• $u' = \frac{d}{dx} (x^2) = 2x$
• $v' = \frac{d}{dx} (x - 1) = 1$

Now, applying the quotient rule:

$f'(x) = \frac{(x - 1) \cdot (2x) - (x^2) \cdot (1)}{(x - 1)^2}$ $f'(x) = \frac{2x(x - 1) - x^2}{(x - 1)^2}$ $f'(x) = \frac{2x^2 - 2x - x^2}{(x - 1)^2}$ $f'(x) = \frac{x^2 - 2x}{(x - 1)^2}$ $f'(x) = \frac{x(x - 2)}{(x - 1)^2}$

Step 2: Set $f'(x) = 0$

To find the points where the function has a horizontal tangent, we set the numerator of the derivative equal to zero:

$x(x - 2) = 0$

This gives us two solutions:

$x = 0 \quad \text{or} \quad x = 2$

Step 3: Determine the corresponding $y$-values

To find the points of the horizontal tangents, substitute $x = 0$ and $x = 2$ back into the original function:

• For $x = 0$: $f(0) = \frac{0^2}{0 - 1} = 0$ So, the point is $(0, 0)$.

• For $x = 2$: $f(2) = \frac{2^2}{2 - 1} = \frac{4}{1} = 4$ So, the point is $(2, 4)$.

Conclusion

The points at which the function has a horizontal tangent are: $(0, 0) \quad \text{and} \quad (2, 4)$

Would you like more details or have any questions?

Here are five related questions for further exploration:

1. How do we identify vertical asymptotes for the function $f(x) = \frac{x^2}{x - 1}$?
2. What is the significance of critical points in determining the shape of a graph?
3. Can you find where the function is increasing or decreasing using $f'(x)$?
4. How does the second derivative help in identifying concavity of the graph?
5. What happens to the function as $x \to 1$ and why?

Tip: Always simplify your derivative as much as possible before solving $f'(x) = 0$ to make your calculations easier!