Question

determine the point at which the graph of the function has a horizontal tangent
f(x)=x/square root of (2x-1)
(x,y)=?

Answer 1

use the quotient rule to get the derivative of the rational
recall that the derivative is the equation of the slope of the tangent to the function

thus when the tangent is just a horizontal line, the slope of a horizontal line is 0
when taking the RISE of it, since it's a horizontal line, both RISE values, 1st and 2nd will be equal to each other and thus \(\bf y_2-y_1=0\)

so set the derivative, or slope function, to 0
solve for "x"

Answer 2

could you show me the set up?

Answer 3

\(\bf f(x)=\cfrac{g(x)}{h(x)}\qquad f'(x)=\cfrac{[g'(x)\cdot h(x)]-[g(x)\cdot h'(x)]}{h(x)^2}\\ \quad \\
f(x)=\cfrac{x}{\sqrt{2x-1}}\\ \quad \\\implies
f'(x)=\cfrac{\left(\frac{d}{dx}[x]\cdot \sqrt{2x-1}\right)-\left(\sqrt{2x-1}\cdot \frac{d}{dx}[\sqrt{2x-1}]\right)}{(\sqrt{2x-1})^2}\)