To find the coordinates of the point on the curve $y = \frac{x^2}{2x - 1}$ where $\frac{dy}{dx} = 0$, let's solve step by step:

Step 1: Differentiate $y$ with respect to $x$.

Given: $y = \frac{x^2}{2x - 1}$ Use the quotient rule for differentiation: $\frac{dy}{dx} = \frac{(v \cdot u') - (u \cdot v')}{v^2},$ where $u = x^2$ and $v = 2x - 1$.

• Derivative of $u = x^2$ is $u' = 2x$,
• Derivative of $v = 2x - 1$ is $v' = 2$.

Substitute into the quotient rule: $\frac{dy}{dx} = \frac{(2x - 1)(2x) - (x^2)(2)}{(2x - 1)^2}.$

Simplify the numerator: $\text{Numerator} = (2x - 1)(2x) - 2x^2.$ Expand: $(2x - 1)(2x) = 4x^2 - 2x.$ Thus: $\text{Numerator} = 4x^2 - 2x - 2x^2 = 2x^2 - 2x.$

So: $\frac{dy}{dx} = \frac{2x^2 - 2x}{(2x - 1)^2}.$

Step 2: Set $\frac{dy}{dx} = 0$.

For the derivative to be zero, the numerator must be zero: $2x^2 - 2x = 0.$ Factorize: $2x(x - 1) = 0.$ Thus: $x = 0 \quad \text{or} \quad x = 1.$

Step 3: Find corresponding $y$-values.

Substitute $x = 0$ and $x = 1$ into $y = \frac{x^2}{2x - 1}$:

1. For $x = 0$: $y = \frac{0^2}{2(0) - 1} = \frac{0}{-1} = 0.$ So, $(x, y) = (0, 0)$.

2. For $x = 1$: $y = \frac{1^2}{2(1) - 1} = \frac{1}{2 - 1} = \frac{1}{1} = 1.$ So, $(x, y) = (1, 1)$.

Step 4: Verify the coordinates.

The points where $\frac{dy}{dx} = 0$ are: $(0, 0) \quad \text{and} \quad (1, 1).$

Final Answer:

From the options given:

• (A) $(0, 0)$
• (B) $(1, 1)$
• (C) $(0, 0), (1, 1)$ (Correct Answer)
• (D) $(0, 1), (2, 0)$

The correct choice is (C) $(0, 0), (1, 1)$.

Any questions or need more clarification?

Here are 5 related questions to deepen understanding:

1. How do you apply the quotient rule for derivatives?
2. Why does setting the numerator of $\frac{dy}{dx}$ to zero solve for critical points?
3. How can we determine if the points $(0, 0)$ and $(1, 1)$ are maxima or minima?
4. What happens to $\frac{dy}{dx}$ when the denominator is zero?
5. Can you sketch the curve $y = \frac{x^2}{2x - 1}$ and identify critical points graphically?

Tip: Always check for undefined points (denominator = 0) when dealing with rational functions!