11. [-/1 Points] DETAILS MY NOTES LARCALC12 2.4.080. Determine the point(s) at which the graph of \[ f(x)=\frac{-4 x}{\sqrt{2 x-1}} \] has a horizontal tangent. \[ (x, y)=(\square) \]

To find the points at which the graph of \( f(x) = \frac{-4 x}{\sqrt{2 x - 1}} \) has a horizontal tangent, we first need to derive the function and set its derivative equal to zero. 1. **Find the derivative:** Using the quotient rule, we differentiate \( f(x) \). Let \( u = -4x \) and \( v = \sqrt{2x - 1} \). The derivative \( f'(x) \) is given by: \[ f'(x) = \frac{u'v - uv'}{v^2} = \frac{-4\sqrt{2x - 1} - (-4x)(\frac{1}{2\sqrt{2x - 1} \cdot 2}) \cdot 2}{(2x-1)} \] This simplifies to finding where \( f'(x) = 0 \). 2. **Set the derivative to zero:** After simplification, set \( f'(x) = 0 \) to determine critical points. 3. **Solve for \( x \)**: Upon solving the derivative, find the corresponding \( y \) values for the identified \( x \) values to find points \( (x, y) \). 4. **Plugging values back into \( f(x) \)**: Once \( x \) values are determined, substituting back into the original function \( f(x) \) gives the actual points. Thus, the points at which the graph of \( f(x) \) has a horizontal tangent are \( (x, y)=(\text{derived } x, f(\text{derived } x)) \). Please proceed with deriving the specific critical points and plug them into the function to complete your result.