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. \[ (x, y)=(\square) \]

We are given   f(x) = (–4x) / √(2x – 1) and we want the point(s) where the graph has a horizontal tangent. A horizontal tangent occurs when f '(x) = 0 (provided the derivative exists). Step 1. Write f(x) in a form that is easier to differentiate:   f(x) = –4x (2x – 1)^(–1/2). Step 2. Differentiate using the product rule:   Let u(x) = –4x  and  v(x) = (2x – 1)^(–1/2). Then u'(x) = –4 and to differentiate v(x) we use the chain rule:   v’(x) = –½ (2x – 1)^(–3/2) · 2 = –(2x – 1)^(–3/2). By the product rule:   f '(x) = u'(x)v(x) + u(x)v’(x)       = –4 (2x – 1)^(–1/2) + (–4x)[–(2x – 1)^(–3/2)]       = –4 (2x – 1)^(–1/2) + 4x (2x – 1)^(–3/2). Step 3. Factor the expression: Notice that (2x – 1)^(–3/2) is a common factor if you write:   f '(x) = (2x – 1)^(–3/2) [–4(2x – 1) + 4x]       = (2x – 1)^(–3/2) [–8x + 4 + 4x]       = (2x – 1)^(–3/2) [–4x + 4]       = 4(1 – x)(2x – 1)^(–3/2). Step 4. Set f '(x) equal to zero: Since (2x – 1)^(–3/2) is never zero (for x > ½, which is required so that √(2x – 1) is defined), we only need:   4(1 – x) = 0 ⇒ 1 – x = 0 ⇒ x = 1. Step 5. Find the y-coordinate: Plug x = 1 into f(x):   f(1) = (–4·1) / √(2·1 – 1) = –4 / √(1) = –4. Thus, the graph has a horizontal tangent at (1, –4). Final Answer:   (x, y) = (1, –4).