# Determine the point at which the graph of the function has a horizontal tangent.?`f(x) = x/sqrt(2x - 1)` (x, y) = ?

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A function has a horizontal tangent at a point where f '(x) = 0.

To determine f'(x), apply the quotient formula of derivatives which is `(u/v)' = (v u' - u v')/v^2` .

So,

`f (x) = x/(2x-1)^(1/2)`

`f'(x) = ((2x-1)^(1/2)*1 - x (1/2)(2x-1)^(-1/2) *2)/((2x-1)^(1/2))^2`

`f'(x) = ((2x-1)^(1/2) - x/(2x-1)^(1/2))/(2x-1)`

To simplify, multiply the top and bottom of the complex fraction by `(2x-1)^(1/2)` .

`f'(x) = (((2x-1)^(1/2)-x/(2x-1)^(1/2)))/((2x-1)) * ((2x-1)^(1/2)/1)/((2x-1)^(1/2)/1)`

`f'(x) = (2x-1 -x)/(2x-1)^(3/2)`

`f'(x) = (x-1)/(2x-1)^(3/2)`

Then, set f'(x) equal to zero. And solve for x.

`0=(x-1)/(2x-1)^(3/2)`

To simplify the equation. multiply both sides by `(2x-1)^(3/2)` .

`0*(2x-1)^(3/2) = (x-1)/(2x-1)^(3/2)*(2x-1)^(3/2)`

`0=x-1`

`1=x`

Substitute the value of x to f(x) to determine y.

`y=f(x) = x/sqrt(2x-1)= 1/sqrt(2*1-1)= 1/sqrt1`

`y=1`

**Hence, the given function has a horizontal tangent at point (1,1).**