Answer
At one point: $(1, 1).$
Work Step by Step
Using the quotient rule: $fâ(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=x; u'(x)=1$
$v(x)=\sqrt{2x-1}; v'(x)=\dfrac{1}{\sqrt{2x-1}}$
$f'(x)=\dfrac{\sqrt{2x-1}-\dfrac{x}{2x-1}}{2x-1}$
$=\dfrac{x-1}{\sqrt{(2x-1)^3}}$
$f'(x)=0 \rightarrow \dfrac{x-1}{\sqrt{(2x-1)^3}}=0 \rightarrow x-1=0 \rightarrow x=1$
$f(1)=\dfrac{1}{\sqrt{2(1)-1}}=1$
Hence the point at which there is a horizontal tangent is $(1, 1).$
